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INTRATIONS SOLUTIONS AND DIFFERENTIATIONAL EQUATIONS

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INTRATIONS SOLUTIONS AND DIFFERENTIATIONAL EQUATIONS
Question 1: Product and Chain Rule
Differentiate f(x)=x3sin (2x2+1)f(x) = x^3 \sin(2x^2 + 1)f(x)=x3sin(2x2+1).
Solution:
To differentiate f(x)=x3sin (2x2+1)f(x) = x^3 \sin(2x^2 + 1)f(x)=x3sin(2x2+1), we will use
both the product rule and the chain rule.
First, recall the product rule:
ddx[u(x)v(x)]=u′(x)v(x)+u(x)v′(x)\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)dxd
[u(x)v(x)]=u′(x)v(x)+u(x)v′(x)
Here, let u(x)=x3u(x) = x^3u(x)=x3 and v(x)=sin (2x2+1)v(x) = \sin(2x^2 +
1)v(x)=sin(2x2+1).
Step 1: Differentiate u(x)u(x)u(x):
u′(x)=ddx[x3]=3x2u'(x) = \frac{d}{dx}[x^3] = 3x^2u′(x)=dxd[x3]=3x2
Step 2: Differentiate v(x)v(x)v(x) using the chain rule:
v(x)=sin (2x2+1)v(x) = \sin(2x^2 + 1)v(x)=sin(2x2+1)
Let g(x)=2x2+1g(x) = 2x^2 + 1g(x)=2x2+1, then v(x)=sin (g(x))v(x) = \
sin(g(x))v(x)=sin(g(x)).
First, find g′(x)g'(x)g′(x):
g′(x)=ddx[2x2+1]=4xg'(x) = \frac{d}{dx}[2x^2 + 1] = 4xg′(x)=dxd[2x2+1]=4x
Next, differentiate sin (g(x))\sin(g(x))sin(g(x)):
ddx[sin (g(x))]=cos (g(x))⋅g′(x)=cos (2x2+1)⋅4x\frac{d}{dx}[\sin(g(x))] = \cos(g(x)) \cdot g'(x)
= \cos(2x^2 + 1) \cdot 4xdxd[sin(g(x))]=cos(g(x))⋅g′(x)=cos(2x2+1)⋅4x
So, v′(x)=4xcos (2x2+1)v'(x) = 4x \cos(2x^2 + 1)v′(x)=4xcos(2x2+1).
Step 3: Apply the product rule:
f′(x)=u′(x)v(x)+u(x)v′(x)f'(x) = u'(x)v(x) + u(x)v'(x)f′(x)=u′(x)v(x)+u(x)v′(x) f′
(x)=3x2sin (2x2+1)+x3⋅4xcos (2x2+1)f'(x) = 3x^2 \sin(2x^2 + 1) + x^3 \cdot 4x \cos(2x^2 +
1)f′(x)=3x2sin(2x2+1)+x3⋅4xcos(2x2+1)
Simplify:
f′(x)=3x2sin (2x2+1)+4x4cos (2x2+1)f'(x) = 3x^2 \sin(2x^2 + 1) + 4x^4 \cos(2x^2 + 1)f′
(x)=3x2sin(2x2+1)+4x4cos(2x2+1)
Question 2: Implicit Differentiation
Given the equation x2+y2=exyx^2 + y^2 = e^{xy}x2+y2=exy, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Differentiate both sides with respect to xxx:
ddx[x2+y2]=ddx[exy]\frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[e^{xy}]dxd[x2+y2]=dxd[exy]
Step 1: Differentiate the left side:
ddx[x2+y2]=2x+2ydydx\frac{d}{dx}[x^2 + y^2] = 2x + 2y \frac{dy}{dx}dxd
[x2+y2]=2x+2ydxdy
Step 2: Differentiate the right side using the chain rule and product rule:
ddx[exy]=exy⋅ddx[xy]=exy⋅(xdydx+y)\frac{d}{dx}[e^{xy}] = e^{xy} \cdot \frac{d}{dx}
[xy] = e^{xy} \cdot (x \frac{dy}{dx} + y)dxd[exy]=exy⋅dxd[xy]=exy⋅(xdxdy+y)
Combine both sides:
2x+2ydydx=exy⋅(xdydx+y)2x + 2y \frac{dy}{dx} = e^{xy} \cdot (x \frac{dy}{dx} +
y)2x+2ydxdy=exy⋅(xdxdy+y)
Step 3: Solve for dydx\frac{dy}{dx}dxdy:
2x+2ydydx=exy⋅xdydx+exy⋅y2x + 2y \frac{dy}{dx} = e^{xy} \cdot x \frac{dy}{dx} +
e^{xy} \cdot y2x+2ydxdy=exy⋅xdxdy+exy⋅y
Collect all dydx\frac{dy}{dx}dxdy terms on one side:
2ydydx−exy⋅xdydx=exy⋅y−2x2y \frac{dy}{dx} - e^{xy} \cdot x \frac{dy}{dx} = e^{xy} \
cdot y - 2x2ydxdy−exy⋅xdxdy=exy⋅y−2x
Factor out dydx\frac{dy}{dx}dxdy:
dydx(2y−exy⋅x)=exy⋅y−2x\frac{dy}{dx} (2y - e^{xy} \cdot x) = e^{xy} \cdot y - 2xdxdy
(2y−exy⋅x)=exy⋅y−2x
Solve for dydx\frac{dy}{dx}dxdy:
dydx=exy⋅y−2x2y−exy⋅x\frac{dy}{dx} = \frac{e^{xy} \cdot y - 2x}{2y - e^{xy} \cdot
x}dxdy=2y−exy⋅xexy⋅y−2x
Question 3: Second Derivative
Find the second derivative of y=ln (cos (x))y = \ln(\cos(x))y=ln(cos(x)).
Solution:
To find the second derivative, we first need to find the first derivative and then differentiate it
again.
Step 1: Find the first derivative dydx\frac{dy}{dx}dxdy.
Let y=ln (cos (x))y = \ln(\cos(x))y=ln(cos(x)).
Using the chain rule:
dydx=ddx[ln (cos (x))]=1cos (x)⋅ddx[cos (x)]\frac{dy}{dx} = \frac{d}{dx}[\ln(\cos(x))] = \
frac{1}{\cos(x)} \cdot \frac{d}{dx}[\cos(x)]dxdy=dxd[ln(cos(x))]=cos(x)1⋅dxd[cos(x)]
Differentiate cos (x)\cos(x)cos(x):
ddx[cos (x)]=−sin (x)\frac{d}{dx}[\cos(x)] = -\sin(x)dxd[cos(x)]=−sin(x)
So, the first derivative is:
dydx=−sin (x)cos (x)=−tan (x)\frac{dy}{dx} = \frac{-\sin(x)}{\cos(x)} = -\tan(x)dxdy=cos(x)
−sin(x)=−tan(x)
Step 2: Find the second derivative d2ydx2\frac{d^2y}{dx^2}dx2d2y.
Differentiate −tan (x)-\tan(x)−tan(x):
d2ydx2=ddx[−tan (x)]=−sec 2(x)\frac{d^2y}{dx^2} = \frac{d}{dx}[-\tan(x)] = -\
sec^2(x)dx2d2y=dxd[−tan(x)]=−sec2(x)
So, the second derivative is:
d2ydx2=−sec 2(x)\frac{d^2y}{dx^2} = -\sec^2(x)dx2d2y=−sec2(x)
Question 4: Differentiation of Parametric Equations
Given the parametric equations x=t2+1x = t^2 + 1x=t2+1 and y=ln (t)y = \ln(t)y=ln(t), find
dydx\frac{dy}{dx}dxdy in terms of ttt.
Solution:
To find dydx\frac{dy}{dx}dxdy in terms of ttt, we need to find dydt\frac{dy}{dt}dtdy and
dxdt\frac{dx}{dt}dtdx, and then use the chain rule.
Step 1: Differentiate x=t2+1x = t^2 + 1x=t2+1 with respect to ttt:
dxdt=2t\frac{dx}{dt} = 2tdtdx=2t
Step 2: Differentiate y=ln (t)y = \ln(t)y=ln(t) with respect to ttt:
dydt=1t\frac{dy}{dt} = \frac{1}{t}dtdy=t1
Step 3: Use the chain rule:
dydx=dydtdxdt=1t2t=12t2\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{\
frac{1}{t}}{2t} = \frac{1}{2t^2}dxdy=dtdxdtdy=2tt1=2t21
Question 5: Differentiation Using Logarithmic Differentiation
Differentiate y=xxy = x^xy=xx.
Solution:
To differentiate y=xxy = x^xy=xx, we will use logarithmic differentiation.
Step 1: Take the natural logarithm of both sides:
ln (y)=ln (xx)\ln(y) = \ln(x^x)ln(y)=ln(xx)
Step 2: Simplify using logarithm properties:
ln (y)=xln (x)\ln(y) = x \ln(x)ln(y)=xln(x)
Step 3: Differentiate both sides with respect to xxx:
ddx[ln (y)]=ddx[xln (x)]\frac{d}{dx}[\ln(y)] = \frac{d}{dx}[x \ln(x)]dxd[ln(y)]=dxd[xln(x)]
Using the chain rule on the left side:
1y⋅dydx=ln (x)+x⋅1x=ln (x)+1\frac{1}{y} \cdot \frac{dy}{dx} = \ln(x) + x \cdot \frac{1}{x}
= \ln(x) + 1y1⋅dxdy=ln(x)+x⋅x1=ln(x)+1
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
dydx=y(ln (x)+1)\frac{dy}{dx} = y (\ln(x) + 1)dxdy=y(ln(x)+1)
Since y=xxy = x^xy=xx:
dydx=xx(ln (x)+1)\frac{dy}{dx} = x^x (\ln(x) + 1)dxdy=xx(ln(x)+1)
These questions and solutions cover a variety of differentiation techniques and provide
detailed steps for solving complex problems.
Question 6: Differentiation Involving Inverse Trigonometric Functions
Find the derivative of y=arctan (x1−x2)y = \arctan\left(\frac{x}{1-x^2}\right)y=arctan(1−x2x
).
Solution:
To differentiate y=arctan (x1−x2)y = \arctan\left(\frac{x}{1-x^2}\right)y=arctan(1−x2x), we
will use the chain rule.
Let u=x1−x2u = \frac{x}{1-x^2}u=1−x2x.
Step 1: Differentiate arctan (u)\arctan(u)arctan(u):
dydu=11+u2\frac{dy}{du} = \frac{1}{1+u^2}dudy=1+u21
Step 2: Differentiate uuu with respect to xxx:
u=x1−x2u = \frac{x}{1-x^2}u=1−x2x
Using the quotient rule:
ddx(x1−x2)=(1−x2)⋅1−x⋅(−2x)(1−x2)2=1−x2+2x2(1−x2)2=1+x2(1−x2)2\frac{d}{dx}\left( \
frac{x}{1-x^2} \right) = \frac{(1-x^2) \cdot 1 - x \cdot (-2x)}{(1-x^2)^2} = \frac{1 - x^2 +
2x^2}{(1-x^2)^2} = \frac{1 + x^2}{(1-x^2)^2}dxd(1−x2x)=(1−x2)2(1−x2)⋅1−x⋅(−2x)
=(1−x2)21−x2+2x2=(1−x2)21+x2
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=11+(x1−x2)2⋅1+x2(1−x2)2\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}
{dx} = \frac{1}{1 + \left( \frac{x}{1-x^2} \right)^2} \cdot \frac{1 + x^2}{(1-x^2)^2}dxdy
=dudy⋅dxdu=1+(1−x2x)21⋅(1−x2)21+x2
Simplify the expression inside the arctangent:
1+(x1−x2)2=1+x2(1−x2)2=(1−x2)2+x2(1−x2)2=1−2x2+x4+x2(1−x2)2=1−x2+x4(1−x2)21 +
\left( \frac{x}{1-x^2} \right)^2 = 1 + \frac{x^2}{(1-x^2)^2} = \frac{(1-x^2)^2 + x^2}{(1-
x^2)^2} = \frac{1 - 2x^2 + x^4 + x^2}{(1-x^2)^2} = \frac{1 - x^2 + x^4}{(1-
x^2)^2}1+(1−x2x)2=1+(1−x2)2x2=(1−x2)2(1−x2)2+x2=(1−x2)21−2x2+x4+x2
=(1−x2)21−x2+x4
This simplifies to:
1+x21−x2\frac{1 + x^2}{1 - x^2}1−x21+x2
Thus, the derivative is:
dydx=11+x21−x2⋅1+x2(1−x2)2=1−x21+x2⋅1+x2(1−x2)2=11−x2\frac{dy}{dx} = \frac{1}{\
frac{1 + x^2}{1 - x^2}} \cdot \frac{1 + x^2}{(1-x^2)^2} = \frac{1 - x^2}{1 + x^2} \cdot \
frac{1 + x^2}{(1-x^2)^2} = \frac{1}{1-x^2}dxdy=1−x21+x21⋅(1−x2)21+x2=1+x21−x2
⋅(1−x2)21+x2=1−x21
Question 7: Differentiation Involving Logarithmic Functions
Differentiate y=ln (x2+1x−1)y = \ln\left(\frac{x^2 + 1}{x - 1}\right)y=ln(x−1x2+1).
Solution:
To differentiate y=ln (x2+1x−1)y = \ln\left(\frac{x^2 + 1}{x - 1}\right)y=ln(x−1x2+1), we
will use the chain rule and the properties of logarithms.
Step 1: Use the properties of logarithms to simplify:
y=ln (x2+1)−ln (x−1)y = \ln(x^2 + 1) - \ln(x - 1)y=ln(x2+1)−ln(x−1)
Step 2: Differentiate each term separately.
First term:
ddx[ln (x2+1)]=1x2+1⋅ddx[x2+1]=1x2+1⋅2x=2xx2+1\frac{d}{dx}[\ln(x^2 + 1)] = \frac{1}
{x^2 + 1} \cdot \frac{d}{dx}[x^2 + 1] = \frac{1}{x^2 + 1} \cdot 2x = \frac{2x}{x^2 +
1}dxd[ln(x2+1)]=x2+11⋅dxd[x2+1]=x2+11⋅2x=x2+12x
Second term:
ddx[ln (x−1)]=1x−1⋅ddx[x−1]=1x−1\frac{d}{dx}[\ln(x - 1)] = \frac{1}{x - 1} \cdot \frac{d}
{dx}[x - 1] = \frac{1}{x - 1}dxd[ln(x−1)]=x−11⋅dxd[x−1]=x−11
Step 3: Combine the results:
dydx=2xx2+1−1x−1\frac{dy}{dx} = \frac{2x}{x^2 + 1} - \frac{1}{x - 1}dxdy=x2+12x
−x−11
Question 8: Differentiation of an Exponential Function
Find the derivative of y=ex2sin (x)y = e^{x^2 \sin(x)}y=ex2sin(x).
Solution:
To differentiate y=ex2sin (x)y = e^{x^2 \sin(x)}y=ex2sin(x), we will use the chain rule.
Let u=x2sin (x)u = x^2 \sin(x)u=x2sin(x).
Step 1: Differentiate eue^ueu:
dydu=eu\frac{dy}{du} = e^ududy=eu
Step 2: Differentiate uuu with respect to xxx:
u=x2sin (x)u = x^2 \sin(x)u=x2sin(x)
Using the product rule:
dudx=ddx[x2]⋅sin (x)+x2⋅ddx[sin (x)]=2xsin (x)+x2cos (x)\frac{du}{dx} = \frac{d}{dx}[x^2] \
cdot \sin(x) + x^2 \cdot \frac{d}{dx}[\sin(x)] = 2x \sin(x) + x^2 \cos(x)dxdu=dxd[x2]⋅sin(x)
+x2⋅dxd[sin(x)]=2xsin(x)+x2cos(x)
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=ex2sin (x)⋅(2xsin (x)+x2cos (x))\frac{dy}{dx} = \frac{dy}{du} \cdot \
frac{du}{dx} = e^{x^2 \sin(x)} \cdot (2x \sin(x) + x^2 \cos(x))dxdy=dudy⋅dxdu
=ex2sin(x)⋅(2xsin(x)+x2cos(x))
Thus, the derivative is:
dydx=ex2sin (x)(2xsin (x)+x2cos (x))\frac{dy}{dx} = e^{x^2 \sin(x)} (2x \sin(x) + x^2 \
cos(x))dxdy=ex2sin(x)(2xsin(x)+x2cos(x))
Question 9: Differentiation Involving Hyperbolic Functions
Differentiate y=sinh −1(x2+1)y = \sinh^{-1}(x^2 + 1)y=sinh−1(x2+1).
Solution:
To differentiate y=sinh −1(x2+1)y = \sinh^{-1}(x^2 + 1)y=sinh−1(x2+1), we will use the
chain rule and the derivative of the inverse hyperbolic sine function.
Step 1: Recall the derivative of sinh −1(u)\sinh^{-1}(u)sinh−1(u):
ddu[sinh −1(u)]=1u2+1\frac{d}{du}[\sinh^{-1}(u)] = \frac{1}{\sqrt{u^2 + 1}}dud
[sinh−1(u)]=u2+11
Step 2: Let u=x2+1u = x^2 + 1u=x2+1. Differentiate uuu with respect to xxx:
dudx=ddx[x2+1]=2x\frac{du}{dx} = \frac{d}{dx}[x^2 + 1] = 2xdxdu=dxd[x2+1]=2x
Step 3: Combine using the chain rule:
dydx=ddu[sinh −1(u)]⋅dudx=1(x2+1)2+1⋅2x=2x(x2+1)2+1\frac{dy}{dx} = \frac{d}{du}[\
sinh^{-1}(u)] \cdot \frac{du}{dx} = \frac{1}{\sqrt{(x^2 + 1)^2 + 1}} \cdot 2x = \frac{2x}{\
sqrt{(x^2 + 1)^2 + 1}}dxdy=dud[sinh−1(u)]⋅dxdu=(x2+1)2+11⋅2x=(x2+1)2+12x
Simplify the expression:
(x2+1)2+1=x4+2x2+1+1=x4+2x2+2(x^2 + 1)^2 + 1 = x^4 + 2x^2 + 1 + 1 = x^4 + 2x^2 +
2(x2+1)2+1=x4+2x2+1+1=x4+2x2+2
Thus, the derivative is:
dydx=2xx4+2x2+2\frac{dy}{dx} = \frac{2x}{\sqrt{x^4 + 2x^2 + 2}}dxdy=x4+2x2+22x
Question 10: Differentiation of a Composite Function
Differentiate y=cos (ln (x2+1))y = \cos(\ln(x^2 + 1))y=cos(ln(x2+1)).
Solution:
To differentiate y=cos (ln (x2+1))y = \cos(\ln(x^2 + 1))y=cos(ln(x2+1)), we will use the chain
rule.
Let u=ln (x2+1)u = \ln(x^2 + 1)u=ln(x2+1) and v=x2+1v = x^2 + 1v=x2+1.
Step 1: Differentiate cos (u)\cos(u)cos(u):
dydu=−sin (u)\frac{dy}{du} = -\sin(u)dudy=−sin(u)
Step 2: Differentiate uuu with respect to vvv:
dudv=1v\frac{du}{dv} = \frac{1}{v}dvdu=v1
Step 3: Differentiate vvv with respect to xxx:
dvdx=2x\frac{dv}{dx} = 2xdxdv=2x
Step 4: Combine using the chain rule:
dydx=dydu⋅dudv⋅dvdx=−sin (ln (x2+1))⋅1x2+1⋅2x\frac{dy}{dx} = \frac{dy}{du} \cdot \
frac{du}{dv} \cdot \frac{dv}{dx} = -\sin(\ln(x^2 + 1)) \cdot \frac{1}{x^2 + 1} \cdot 2xdxdy
=dudy⋅dvdu⋅dxdv=−sin(ln(x2+1))⋅x2+11⋅2x
Thus, the derivative is:
dydx=−2xsin (ln (x2+1))x2+1\frac{dy}{dx} = -\frac{2x \sin(\ln(x^2 + 1))}{x^2 + 1}dxdy
=−x2+12xsin(ln(x2+1))
These additional questions and solutions cover a wide range of differentiation techniques and
applications, providing more practice with complex differentiation problems.
Sure! Here are more advanced differentiation questions along with their detailed solutions:
Question 11: Differentiation of a Function with Nested Trigonometric and
Exponential Functions
Differentiate y=ecos (3x2)y = e^{\cos(3x^2)}y=ecos(3x2).
Solution:
To differentiate y=ecos (3x2)y = e^{\cos(3x^2)}y=ecos(3x2), we will use the chain rule.
Let u=cos (3x2)u = \cos(3x^2)u=cos(3x2).
Step 1: Differentiate eue^ueu:
dydu=eu\frac{dy}{du} = e^ududy=eu
Step 2: Differentiate u=cos (3x2)u = \cos(3x^2)u=cos(3x2) using the chain rule:
dudx=−sin (3x2)⋅ddx[3x2]=−sin (3x2)⋅6x=−6xsin (3x2)\frac{du}{dx} = -\sin(3x^2) \cdot \
frac{d}{dx}[3x^2] = -\sin(3x^2) \cdot 6x = -6x \sin(3x^2)dxdu=−sin(3x2)⋅dxd
[3x2]=−sin(3x2)⋅6x=−6xsin(3x2)
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=ecos (3x2)⋅(−6xsin (3x2))\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}
{dx} = e^{\cos(3x^2)} \cdot (-6x \sin(3x^2))dxdy=dudy⋅dxdu=ecos(3x2)⋅(−6xsin(3x2))
Thus, the derivative is:
dydx=−6xecos (3x2)sin (3x2)\frac{dy}{dx} = -6x e^{\cos(3x^2)} \sin(3x^2)dxdy
=−6xecos(3x2)sin(3x2)
Question 12: Differentiation of a Function with a Logarithmic and
Trigonometric Composition
Differentiate y=ln (tan (x2))y = \ln(\tan(x^2))y=ln(tan(x2)).
Solution:
To differentiate y=ln (tan (x2))y = \ln(\tan(x^2))y=ln(tan(x2)), we will use the chain rule.
Let u=tan (x2)u = \tan(x^2)u=tan(x2).
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate u=tan (x2)u = \tan(x^2)u=tan(x2) using the chain rule:
dudx=sec 2(x2)⋅ddx[x2]=sec 2(x2)⋅2x=2xsec 2(x2)\frac{du}{dx} = \sec^2(x^2) \cdot \frac{d}
{dx}[x^2] = \sec^2(x^2) \cdot 2x = 2x \sec^2(x^2)dxdu=sec2(x2)⋅dxd
[x2]=sec2(x2)⋅2x=2xsec2(x2)
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1tan (x2)⋅2xsec 2(x2)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
= \frac{1}{\tan(x^2)} \cdot 2x \sec^2(x^2)dxdy=dudy⋅dxdu=tan(x2)1⋅2xsec2(x2)
Thus, the derivative is:
dydx=2xsec 2(x2)tan (x2)\frac{dy}{dx} = \frac{2x \sec^2(x^2)}{\tan(x^2)}dxdy
=tan(x2)2xsec2(x2)
Question 13: Differentiation Involving Multiple Functions
Differentiate y=exsin (x)x2+1y = \frac{e^x \sin(x)}{x^2 + 1}y=x2+1exsin(x).
Solution:
To differentiate y=exsin (x)x2+1y = \frac{e^x \sin(x)}{x^2 + 1}y=x2+1exsin(x), we will use
the quotient rule.
Recall the quotient rule:
ddx[u(x)v(x)]=u′(x)v(x)−u(x)v′(x)[v(x)]2\frac{d}{dx}\left[\frac{u(x)}{v(x)}\right] = \
frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}dxd[v(x)u(x)]=[v(x)]2u′(x)v(x)−u(x)v′(x)
Here, let u(x)=exsin (x)u(x) = e^x \sin(x)u(x)=exsin(x) and v(x)=x2+1v(x) = x^2 +
1v(x)=x2+1.
Step 1: Differentiate u(x)u(x)u(x):
u′(x)=ddx[exsin (x)]=exsin (x)+excos (x)=ex(sin (x)+cos (x))u'(x) = \frac{d}{dx}[e^x \sin(x)] =
e^x \sin(x) + e^x \cos(x) = e^x (\sin(x) + \cos(x))u′(x)=dxd[exsin(x)]=exsin(x)
+excos(x)=ex(sin(x)+cos(x))
Step 2: Differentiate v(x)v(x)v(x):
v′(x)=ddx[x2+1]=2xv'(x) = \frac{d}{dx}[x^2 + 1] = 2xv′(x)=dxd[x2+1]=2x
Step 3: Apply the quotient rule:
dydx=ex(sin (x)+cos (x))(x2+1)−exsin (x)⋅2x(x2+1)2\frac{dy}{dx} = \frac{e^x (\sin(x) + \
cos(x))(x^2 + 1) - e^x \sin(x) \cdot 2x}{(x^2 + 1)^2}dxdy=(x2+1)2ex(sin(x)+cos(x))
(x2+1)−exsin(x)⋅2x
Simplify the numerator:
dydx=ex[(sin (x)+cos (x))(x2+1)−2xsin (x)](x2+1)2\frac{dy}{dx} = \frac{e^x \left[(\sin(x) + \
cos(x))(x^2 + 1) - 2x \sin(x)\right]}{(x^2 + 1)^2}dxdy=(x2+1)2ex[(sin(x)+cos(x))
(x2+1)−2xsin(x)]
Distribute exe^xex:
dydx=ex[x2sin (x)+sin (x)+x2cos (x)+cos (x)−2xsin (x)](x2+1)2\frac{dy}{dx} = \frac{e^x
[x^2 \sin(x) + \sin(x) + x^2 \cos(x) + \cos(x) - 2x \sin(x)]}{(x^2 + 1)^2}dxdy
=(x2+1)2ex[x2sin(x)+sin(x)+x2cos(x)+cos(x)−2xsin(x)]
Thus, the derivative is:
dydx=ex[x2(sin (x)+cos (x))+sin (x)+cos (x)−2xsin (x)](x2+1)2\frac{dy}{dx} = \frac{e^x [x^2
(\sin(x) + \cos(x)) + \sin(x) + \cos(x) - 2x \sin(x)]}{(x^2 + 1)^2}dxdy=(x2+1)2ex[x2(sin(x)
+cos(x))+sin(x)+cos(x)−2xsin(x)]
Question 14: Differentiation of a Function Involving Inverse Trigonometric
and Exponential Functions
Differentiate y=arcsin (ex)y = \arcsin(e^x)y=arcsin(ex).
Solution:
To differentiate y=arcsin (ex)y = \arcsin(e^x)y=arcsin(ex), we will use the chain rule.
Let u=exu = e^xu=ex.
Step 1: Differentiate arcsin (u)\arcsin(u)arcsin(u):
dydu=11−u2\frac{dy}{du} = \frac{1}{\sqrt{1 - u^2}}dudy=1−u21
Step 2: Differentiate u=exu = e^xu=ex:
dudx=ex\frac{du}{dx} = e^xdxdu=ex
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=11−(ex)2⋅ex=ex1−e2x\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}
= \frac{1}{\sqrt{1 - (e^x)^2}} \cdot e^x = \frac{e^x}{\sqrt{1 - e^{2x}}}dxdy=dudy⋅dxdu
=1−(ex)21⋅ex=1−e2xex
Thus, the derivative is:
dydx=ex1−e2x\frac{dy}{dx} = \frac{e^x}{\sqrt{1 - e^{2x}}}dxdy=1−e2xex
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
xdydx(1−y)=xy−yx \frac{dy}{dx} (1 - y) = xy - yxdxdy(1−y)=xy−y dydx(1−y)=xy−yx\
frac{dy}{dx} (1 - y) = \frac{xy - y}{x}dxdy(1−y)=xxy−y dydx=y1−y\frac{dy}{dx} = \
frac{y}{1 - y}dxdy=1−yy
Thus, the derivative is:
dydx=y1−y\frac{dy}{dx} = \frac{y}{1 - y}dxdy=1−yy
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
Question 15: Differentiation of a Composite Function Involving Logarithms
and Polynomials
Differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5).
Solution:
To differentiate y=ln (x3+3x+5)y = \ln(x^3 + 3x + 5)y=ln(x3+3x+5), we will use the chain
rule.
Let u=x3+3x+5u = x^3 + 3x + 5u=x3+3x+5.
Step 1: Differentiate ln (u)\ln(u)ln(u):
dydu=1u\frac{dy}{du} = \frac{1}{u}dudy=u1
Step 2: Differentiate uuu with respect to xxx:
dudx=ddx[x3+3x+5]=3x2+3\frac{du}{dx} = \frac{d}{dx}[x^3 + 3x + 5] = 3x^2 + 3dxdu
=dxd[x3+3x+5]=3x2+3
Step 3: Combine using the chain rule:
dydx=dydu⋅dudx=1x3+3x+5⋅(3x2+3)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \
frac{1}{x^3 + 3x + 5} \cdot (3x^2 + 3)dxdy=dudy⋅dxdu=x3+3x+51⋅(3x2+3)
Thus, the derivative is:
dydx=3(x2+1)x3+3x+5\frac{dy}{dx} = \frac{3(x^2 + 1)}{x^3 + 3x + 5}dxdy
=x3+3x+53(x2+1)
Question 16: Differentiation Involving Implicit Differentiation and
Logarithms
Given the equation ln (xy)=x+y\ln(xy) = x + yln(xy)=x+y, find dydx\frac{dy}{dx}dxdy.
Solution:
To find dydx\frac{dy}{dx}dxdy, we will use implicit differentiation.
Step 1: Differentiate both sides with respect to xxx:
ddx[ln (xy)]=ddx[x+y]\frac{d}{dx}[\ln(xy)] = \frac{d}{dx}[x + y]dxd[ln(xy)]=dxd[x+y]
Using the chain rule on the left side:
1xy⋅(y+xdydx)=1+dydx\frac{1}{xy} \cdot \left( y + x \frac{dy}{dx} \right) = 1 + \frac{dy}
{dx}xy1⋅(y+xdxdy)=1+dxdy
Step 2: Simplify the equation:
y+xdydxxy=1+dydx\frac{y + x \frac{dy}{dx}}{xy} = 1 + \frac{dy}{dx}xyy+xdxdy=1+dxdy
Multiply both sides by xyxyxy:
y+xdydx=xy+xydydxy + x \frac{dy}{dx} = xy + xy \frac{dy}{dx}y+xdxdy=xy+xydxdy
Step 3: Collect all dydx\frac{dy}{dx}dxdy terms on one side:
y+xdydx−xydydx=xyy + x \frac{dy}{dx} - xy \frac{dy}{dx} = xyy+xdxdy−xydxdy=xy
y+xdydx(1−y)=xyy + x \frac{dy}{dx} (1 - y) = xyy+xdxdy(1−y)=xy
Step 4: Solve for dydx\frac{dy}{dx}dxdy:
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