MATH 108 - ELEMENTARY AND INTERMEDIATE ALGEBRA Differentiation of Trigonometric Functions Question Bank Set 2 | MATH 108

MATH 108 - ELEMENTARY AND

INTERMEDIATE ALGEBRA -

Differentiation of Trigonometric

Functions

Question Bank - Set 2

Liberty University

Question 1

Question

Find the derivative of the function f(x) = cos2(x) sin3(x).

Solution

To find the derivative of the given function, we will use the product rule and

the chain rule of differentiation.

Step 1: Apply the product rule. Let u= cos2(x) and v= sin3(x). Then,

the function f(x) can be expressed as f(x) = u·v.

The product rule states that if f(x) = u·v, then f′(x) = u′·v+u·v′.

Step 2: Find du

dx and dv

dx . Using the chain rule and power rule, we have:

u= cos2(x) =⇒u′= 2 cos(x)(−sin(x)) = −2 cos(x) sin(x)

v= sin3(x) =⇒v′= 3 sin2(x) cos(x)

Step 3: Apply the product rule formula. Now, we substitute u′,v′into the

product rule formula:

f′(x) = −2 cos(x) sin(x)·sin3(x) + cos2(x)·3 sin2(x) cos(x)

Step 4: Simplify the expression. Simplify the expression to get the final

derivative:

f′(x) = −2 cos(x) sin(x) sin3(x) + 3 cos2(x) sin2(x) cos(x)

Therefore, the derivative of f(x) = cos2(x) sin3(x) is f′(x) = −2 cos(x) sin(x) sin3(x)+

3 cos2(x) sin2(x) cos(x).

Question 2

Question

Find the derivative of the function f(x) = cos2(3x) + sin(2x).

Solution

To find the derivative of f(x), we will use the chain rule and the sum rule for

differentiation.

Step 1: Find the derivative of cos2(3x) using the chain rule.

dx (cos2(3x)) = 2 cos(3x)·d

dx (cos(3x))

= 2 cos(3x)(−3 sin(3x))

=−6 cos(3x) sin(3x)

Step 2: Find the derivative of sin(2x).

dx (sin(2x)) = 2 cos(2x)

Step 3: Putting it all together, find the derivative of f(x).

f′(x) = d

dx (cos2(3x) + sin(2x))

=−6 cos(3x) sin(3x) + 2 cos(2x)

Question 3

Question

Find the derivative of the function f(x) = √3 cos(x)−sin(x)

cos(x).

Solution

Step 1: Let’s simplify the given function before finding the derivative. Step 2:

We can simplify f(x) by multiplying the numerator and denominator by cos(x).

Step 3: After simplifying, the function becomes f(x) = √3−tan(x). Step 4:

Now, we can find the derivative of f(x). Step 5: Recall that the derivative of a

constant is zero, and the derivative of tan(x) is sec2(x). Step 6: Therefore, the

derivative of f(x) is f′(x)=0−sec2(x). Step 7: Simplifying further, we get

f′(x) = −sec2(x).

Question 4

Question

Find the derivative of the function f(x) = sin(x) cos(x) with respect to x.

Solution

To find the derivative of f(x) = sin(x) cos(x), we will use the product rule for

differentiation.

Step 1: Apply the product rule, which states that if u(x) and v(x) are

differentiable functions, then the derivative of u(x)v(x) is u′(x)v(x)+u(x)v′(x).

Let u(x) = sin(x) and v(x) = cos(x). Then, u′(x) = cos(x) and v′(x) =

−sin(x).

Step 2: Now, apply the product rule to find the derivative of f(x):

f′(x) = u′(x)v(x) + u(x)v′(x)

= cos(x)·cos(x) + sin(x)·(−sin(x))

= cos2(x)−sin2(x)

= cos(2x).

Therefore, the derivative of f(x) = sin(x) cos(x) with respect to xis f′(x) =

cos(2x).

Question 5

Question

Find the derivative of the function f(x) = sin2(x) cos(x).

Solution

Step 1: To find the derivative of f(x), we will use the product rule. Step 2: The

product rule states that if f(x) = u(x)v(x), then f′(x) = u′(x)v(x) + u(x)v′(x).

Step 3: Let u(x) = sin2(x) and v(x) = cos(x). Step 4: Find the derivatives

of u(x) and v(x). Step 5: u′(x) = 2 sin(x) cos(x) by the chain rule. Step

6: v′(x) = −sin(x). Step 7: Apply the product rule to find f′(x). Step 8:

f′(x) = (2 sin(x) cos(x)·cos(x)) + (sin2(x)· −sin(x)). Step 9: Simplify the

expression. Step 10: f′(x) = 2 sin(x) cos2(x)−sin3(x). Step 11: Therefore, the

derivative of f(x) = sin2(x) cos(x) is f′(x) = 2 sin(x) cos2(x)−sin3(x).

Question 6

Question

Differentiate the function f(x) = sin2(x) cos(x) with respect to x.

Solution

Step 1: Apply the product rule to differentiate the function.

Step 2: Let u= sin2(x) and v= cos(x).

Step 3: Compute u′and v′.

u′=d

dx (sin2(x))

= 2 sin(x) cos(x)

v′=d

dx (cos(x))

=−sin(x)

Step 4: Apply the product rule: f′(x) = u′v+uv′.

f′(x) = (2 sin(x) cos(x))(cos(x)) + (sin2(x))(−sin(x))

= 2 sin(x) cos2(x)−sin3(x)

Therefore, the derivative of f(x) = sin2(x) cos(x) with respect to xis f′(x) =

2 sin(x) cos2(x)−sin3(x).

Question 7

Question

Find the derivative of the function f(x) = sin3x2−2.

Solution

Step 1: Apply the chain rule, where the derivative of sin(u) is cos(u)·u′. Step

2: Identify uas the function inside the sine function, namely u= 3x2−2. Step

3: Compute u′, the derivative of uwith respect to x. Step 4: Substitute uand

u′into the chain rule formula. Step 5: Simplify the expression to get the final

derivative.

Step 1: Apply the chain rule:

dx (sin(u)) = cos(u)·u′

Step 2: Identify uas 3x2−2.

Step 3: Compute u′:

dx = 6x

Step 4: Substitute uand u′into the chain rule formula:

dx (sin3x2−2) = cos3x2−2·6x

Step 5: Simplify the expression:

dx (sin3x2−2)=6xcos3x2−2

Therefore, the derivative of the function f(x) = sin3x2−2is 6xcos3x2−2.

Question 8

Question

Find the derivative of the function f(x) = sin2(x) cos2(x).

Solution

Step 1: We can start by using the product rule for differentiation, which states

that if f(x) = u(x)v(x), then f′(x) = u′(x)v(x) + u(x)v′(x).

Step 2: Let u(x) = sin2(x) and v(x) = cos2(x). Then we have u′(x) =

2 sin(x) cos(x) and v′(x) = −2 cos(x) sin(x).

Step 3: Applying the product rule, we get

f′(x) = u′(x)v(x) + u(x)v′(x)

= (2 sin(x) cos(x))(cos2(x)) + (sin2(x))(−2 cos(x) sin(x))

= 2 sin(x) cos(x) cos2(x)−2 sin2(x) cos(x) sin(x).

Step 4: Simplifying further,

f′(x) = 2 sin(x) cos3(x)−2 sin3(x) cos(x).

Therefore, the derivative of the function f(x) = sin2(x) cos2(x) is f′(x) =

2 sin(x) cos3(x)−2 sin3(x) cos(x).

Question 9

Question

Find the derivative of the function f(x) = sin(x)2cos(x).

Solution

Step 1: Apply the product rule, (uv)′=u′v+uv′, where u= sin(x)2and

v= cos(x). Step 2: Find u′and v′using the chain rule and product rule,

respectively.

u′=d

dx (sin(x)2)

= 2 sin(x) cos(x) (using the chain rule)

v′=d

dx (cos(x))

=−sin(x) (using the derivative of cosine function)

Step 3: Now, apply the product rule to find f′(x).

f′(x)=(u′v+uv′)

= (2 sin(x) cos(x))(cos(x)) + (sin(x)2)(−sin(x))

= 2 sin(x) cos(x)2−sin(x)3cos(x)

Therefore, the derivative of the function f(x) = sin(x)2cos(x) is f′(x) =

2 sin(x) cos(x)2−sin(x)3cos(x).

Question 10

Question

Find the derivative of the function f(x) = cos2(x) sin(x).

Solution

Step 1: Apply the product rule to differentiate f(x):

f′(x) = d

dx (cos2(x) sin(x))

dx (cos2(x)) ·sin(x) + cos2(x)d

dx (sin(x))

Step 2: Differentiate cos2(x) with respect to xusing the chain rule:

dx (cos2(x)) = 2 cos(x)·d

dx (cos(x))

= 2 cos(x)·(−sin(x))

Step 3: Differentiate sin(x) with respect to x:

dx (sin(x)) = cos(x)

Step 4: Substitute the derivatives back into f′(x):

f′(x) = (2 cos(x)·(−sin(x))) ·sin(x) + cos2(x)·cos(x)

=−2 cos(x) sin2(x) + cos3(x)

Therefore, the derivative of f(x) = cos2(x) sin(x) is f′(x) = −2 cos(x) sin2(x)+

cos3(x).

Question 11

Question

Find the derivative of the function f(x) = sin2(2x) + cos2(3x).

Solution

Step 1: Recall the trigonometric identity sin2(θ) + cos2(θ) = 1 for any angle θ.

Step 2: Differentiate f(x) term by term using the chain rule and the trigono-

metric identities. Step 3: Let’s differentiate f(x) = sin2(2x) + cos2(3x) step by

step.

dx [sin2(2x)] = 2 sin(2x)·cos(2x) (Applying chain rule)

= 2 sin(2x)·cos(2x)

= sin(4x) (Using double angle formula for sine)

dx [cos2(3x)] = −2 cos(3x)·sin(3x) (Applying chain rule)

=−2 cos(3x)·sin(3x)

=−sin(6x) (Using double angle formula for cosine)

Step 4: Add the derivatives of the individual terms to find the derivative of

f(x).

dx [f(x)] = d

dx [sin2(2x)] + d

dx [cos2(3x)]

= sin(4x)−sin(6x)

= sin(4x) + sin(−6x)

= sin(4x)−sin(6x) (Since sin is an odd function)

Therefore, the derivative of the function f(x) = sin2(2x) + cos2(3x) is f′(x) =

sin(4x)−sin(6x).

Question 12

Question

Find the derivative of f(x) = sin(x)

1+cos(x).

Solution

Step 1: To differentiate f(x), we will first simplify it using trigonometric iden-

tities. Step 2: Write f(x) as sin(x)

1+cos(x)=sin(x)(1−cos(x))

1−cos2(x). Step 3: Recall the

Pythagorean identity sin2(x) + cos2(x) = 1. Step 4: This allows us to rewrite

sin(x)(1−cos(x))

1−cos2(x)as sin(x)(1−cos(x))

sin2(x). Step 5: Simplify to get f(x) = sin(x)−sin(x) cos(x)

sin2(x).

Step 6: Now, differentiate f(x) using the quotient rule u

v′=u′v−uv′

v2. Step 7:

Let u= sin(x)−sin(x) cos(x) and v= sin(x). Compute the derivatives u′and

v′. Step 8: We have u′= (cos(x)−cos2(x)) = cos(x)−cos2(x) and v′= cos(x).

Step 9: Apply the quotient rule to find f′(x):

f′(x) = (cos(x)−cos2(x)) sin(x)−(sin(x)−sin(x) cos(x)) cos(x)

sin2(x)

Step 10: Simplify f′(x) to get the final answer. Step 11: After simplification,

f′(x) is cos(x) sin(x)−cos2(x) sin(x)−sin(x) cos(x)+sin2(x) cos(x)

sin2(x). Step 12: Simplify fur-

ther to get f′(x) = sin(x)−sin2(x)−sin(x) cos(x)+sin2(x) cos(x)

sin2(x). Step 13: Finally, sim-

plify f′(x) to get f′(x) = sin(x)−sin(x) cos(x)

sin2(x). Step 14: Therefore, the derivative

of f(x) = sin(x)

1+cos(x)is f′(x) = sin(x)−sin(x) cos(x)

sin2(x).

Question 13

Question

Find the derivative of the function f(x) = sin3(x) + cos2(x).

Solution

Step 1: To find the derivative of f(x), we will differentiate each term separately

using the power rule and chain rule where necessary. Step 2: Let’s first find the

derivative of sin3(x). Step 3: We have d

dx (sin3(x)) = 3 sin2(x) cos(x) by applying

the chain rule. Step 4: Now, let’s find the derivative of cos2(x). Step 5: We have

dx (cos2(x)) = 2 cos(x)(−sin(x)) by applying the chain rule. Step 6: Combining

the derivatives of the two terms, we get d

dx (sin3(x)+cos2(x)) = 3 sin2(x) cos(x)+

2 cos(x)(−sin(x)). Step 7: Simplifying this expression, we obtain the derivative

as 3 sin2(x) cos(x)−2 cos(x) sin(x). Therefore, the derivative of the function

f(x) = sin3(x) + cos2(x) is 3 sin2(x) cos(x)−2 cos(x) sin(x).

Question 14

Question

Find the derivative of the function f(x) = sin(3x) cos(2x).

Solution

To find the derivative of f(x) = sin(3x) cos(2x), we will use the product rule of

differentiation which states that if f(x) = u(x)v(x), then f′(x) = u′(x)v(x) +

u(x)v′(x).

Step 1: Let u(x) = sin(3x) and v(x) = cos(2x).

Step 2: Find u′(x) and v′(x).



The derivative of sin(3x) with respect to xis 3 cos(3x), using the chain

rule.



The derivative of cos(2x) with respect to xis −2 sin(2x), using the chain

rule.

Step 3: Apply the product rule.

f′(x) = u′(x)v(x) + u(x)v′(x) = (3 cos(3x))(cos(2x)) + (sin(3x))(−2 sin(2x))

Therefore, the derivative of the function f(x) = sin(3x) cos(2x) is f′(x) =

3 cos(3x) cos(2x)−2 sin(3x) sin(2x).

Question 15

Question

Find the derivative of y= sin(3x) cos(2x) with respect to x.

Solution

Step 1: Apply the product rule to differentiate ywith respect to x. Step 2: Let

u= sin(3x) and v= cos(2x). Step 3: Find du

dx and dv

dx . Step 4: Use the product

rule formula d

dx (uv) = udv

dx +vdu

dx . Step 5: Substitute u,v,du

dx , and dv

dx into the

product rule formula and simplify to find the derivative of y.

Question 16

Question

Find the derivative of the function f(x) = cos(x) sin(x)

x2.

Solution

To find the derivative of the given function, we will use the quotient rule and

the product rule for differentiation.

f(x) = cos(x) sin(x)

= cos(x)·sin(x)

= cos(x)·sin(x)·x−2

Step 1: Apply the product rule to find f′(x).

f′(x) = (cos(x))′·sin(x)·x−2+ cos(x)·(sin(x))′·x−2+ cos(x)·sin(x)·(−2x−3)

= (−sin(x)) ·sin(x)·x−2+ cos(x)·cos(x)·x−2−2 cos(x)·sin(x)·x−3

f′(x) = −sin2(x)·x−2+ cos2(x)·x−2−2 cos(x) sin(x)·x−3

Step 2: Simplify the derivative.

f′(x) = cos2(x)−sin2(x)

x2−2 sin(x) cos(x)

Thus, the derivative of the function f(x) = cos(x) sin(x)

x2is f′(x) = cos2(x)−sin2(x)

x2−

2 sin(x) cos(x)

x3.

Question 17

Question

Find the derivative of the function f(x) = sin(2x) cos(3x).

Solution

Step 1: Apply the product rule to differentiate f(x). Step 2: Let u= sin(2x)

and v= cos(3x). Step 3: Calculate u′and v′. Step 4: Apply the product rule

formula: (uv)′=u′v+uv′. Step 5: Substitute u,v,u′, and v′back into the

formula to find f′(x).

Therefore, the derivative of the function f(x) = sin(2x) cos(3x) is f′(x) =

2 cos(2x) cos(3x)−3 sin(2x) sin(3x).

Question 18

Question

Find the derivative of the function y= sin2(3x) + cos2(2x).

Solution

Step 1: Apply the chain rule to differentiate sin2(3x) with respect to x.

Let u= sin(3x) =⇒du

dx = 3 cos(3x).

dx (sin2(3x)) = 2 sin(3x)·du

dx = 2 sin(3x)·3 cos(3x) = 6 sin(3x) cos(3x).

Step 2: Apply the chain rule to differentiate cos2(2x) with respect to x.

Let v= cos(2x) =⇒dv

dx =−2 sin(2x).

dx (cos2(2x)) = 2 cos(2x)·dv

dx = 2 cos(2x)· −2 sin(2x) = −4 cos(2x) sin(2x).

Step 3: Add the derivatives found in Step 1 and Step 2 to get the final

derivative of the function y= sin2(3x) + cos2(2x).

y′= 6 sin(3x) cos(3x)−4 cos(2x) sin(2x).

Question 19

Question

Find the derivative of f(x) = cos(2x) tan(3x).

Solution

Step 1: Use the product rule to differentiate f(x) = cos(2x) tan(3x). Step 2:

Let u= cos(2x) and v= tan(3x). Step 3: Find u′and v′. Step 4:

u′=−sin(2x)·2

=−2 sin(2x).

Step 5:

v′= sec2(3x)·3

= 3 sec2(3x).

Step 6: Apply the product rule f′(x) = u′v+uv′. Step 7:

f′(x)=(−2 sin(2x)) tan(3x) + cos(2x)(3 sec2(3x))

=−2 sin(2x) tan(3x) + 3 cos(2x) sec2(3x).

Therefore, the derivative of f(x) = cos(2x) tan(3x) is −2 sin(2x) tan(3x) +

3 cos(2x) sec2(3x).

Question 20

Question

Differentiate the function f(x) = sin2(3x) + cos2(4x) with respect to x.

Solution

Step 1: Recall the trigonometric identities sin2(θ)+cos2(θ) = 1 and d

dθ (sin(θ)) =

cos(θ) and d

dθ (cos(θ)) = −sin(θ).

Step 2: Rewrite the given function using the trigonometric identity: f(x) =

Step 3: Differentiate f(x) with respect to x:d

dx f(x) = d

dx 1.

Step 4: The derivative of a constant is zero, so d

dx f(x) = 0.

Therefore, the derivative of f(x) = sin2(3x) + cos2(4x) with respect to xis

0 .

Question 21

Question

Find the derivative of the function f(x) = sin2(x) cos(x).

Solution

To find the derivative of f(x), we will use the product rule. Recall that the

product rule states that if uand vare differentiable functions of x, then the

derivative of their product is given by:

(uv)′=u′v+uv′

Step 1: Let u= sin2(x) and v= cos(x).

Step 2: Find u′and v′.

Using the chain rule, we have:

u′= 2 sin(x) cos(x)

And:

v′=−sin(x)

Step 3: Apply the product rule to find f′(x).

f′(x) = (u′v)+(uv′)

= (2 sin(x) cos(x)·cos(x)) + (sin2(x)· −sin(x))

= 2 sin(x) cos2(x)−sin3(x)

Therefore, the derivative of the function f(x) = sin2(x) cos(x) is f′(x) =

2 sin(x) cos2(x)−sin3(x).

Question 22

Question

Find the derivative of the function f(x) = sin(x) cos(x).

Solution

Step 1: Apply the product rule, which states that if f(x) = g(x)·h(x), then

f′(x) = g′(x)h(x) + g(x)h′(x).

Let g(x) = sin(x) and h(x) = cos(x). Then, g′(x) = cos(x) and h′(x) =

−sin(x).

Step 2: Substitute into the product rule formula.

f′(x) = (sin(x))(−sin(x)) + (cos(x))(cos(x))

Step 3: Simplify the expression.

f′(x) = −sin2(x) + cos2(x)

Step 4: Remember the trigonometric identity sin2(x) + cos2(x) = 1.

Step 5: Rewrite the expression using the trigonometric identity.

f′(x) = −1 + 1 = 0

Therefore, the derivative of the function f(x) = sin(x) cos(x) is 0.

Question 23

Question

Find the derivative of the function f(x) = sin2(2x)−cos2(3x).

Solution

To find the derivative of f(x), we will differentiate each term separately using

the chain rule and trigonometric identities.

Step 1: Find the derivative of sin2(2x).

dx (sin2(2x)) = 2 sin(2x) cos(2x)·2 = 4 sin(2x) cos(2x)

Step 2: Find the derivative of cos2(3x).

dx (cos2(3x)) = −2 cos(3x) sin(3x)·3 = −6 cos(3x) sin(3x)

Step 3: Combine the derivatives to find d

dx (f(x)).

dx (f(x)) = 4 sin(2x) cos(2x)+(−6 cos(3x) sin(3x))

dx (f(x)) = 2 sin(4x)−3 sin(6x)

Therefore, the derivative of f(x) = sin2(2x)−cos2(3x) is d

dx (f(x)) = 2 sin(4x)−

3 sin(6x).

Question 24

Question

Find the derivative of y= sin(2x) cos(3x) with respect to x.

Solution

Step 1: Apply the product rule to differentiate y= sin(2x) cos(3x).

Step 2: Let u= sin(2x) and v= cos(3x). Then, using the product rule

(uv)′=u′v+uv′, we have:

y′= (u)′v+u(v)′

= (2 cos(2x)) cos(3x) + sin(2x) (−3 sin(3x))

= 2 cos(2x) cos(3x)−3 sin(2x) sin(3x).

Therefore, the derivative of y= sin(2x) cos(3x) with respect to xis 2 cos(2x) cos(3x)−

3 sin(2x) sin(3x).

Question 25

Question

Find the derivative of the function f(x) = tan(x) + sec(x).

Solution

Step 1: To find the derivative of f(x), we will use the trigonometric identities

dx (tan(x)) = sec2(x) and d

dx (sec(x)) = sec(x) tan(x).

Step 2: Let’s find d

dx (tan(x)) first. Using the derivative of tan(x) identity,

we have: d

dx (tan(x)) = sec2(x).

Step 3: Next, let’s find d

dx (sec(x)). Using the derivative of sec(x) identity,

we have: d

dx (sec(x)) = sec(x) tan(x).

Step 4: Now, let’s find the derivative of the function f(x) by adding the

derivatives of tan(x) and sec(x):

dx (f(x)) = d

dx (tan(x)) + d

dx (sec(x)).

Step 5: Substitute the derivative expressions we found earlier:

dx (f(x)) = sec2(x) + sec(x) tan(x).

Therefore, the derivative of the function f(x) = tan(x) + sec(x) is f′(x) =

sec2(x) + sec(x) tan(x).

Question 26

Question

Find the derivative of y= sin2(3x) + cos2(3x) with respect to x.

Solution

Step 1: Use the trigonometric identity sin2θ+ cos2θ= 1.

sin2(3x) + cos2(3x)=1

Step 2: Differentiate both sides of the equation with respect to x.

dx (sin2(3x) + cos2(3x)) = d

dx (1)

Step 3: Apply the sum rule and the chain rule to differentiate the left side

of the equation.

dx (sin2(3x)) + d

dx (cos2(3x)) = 0

Step 4: Apply the chain rule to differentiate sin2(3x) and cos2(3x).

2 sin(3x) cos(3x)·3 + 2 cos(3x)(−sin(3x)·3) = 0

Step 5: Simplify the expression.

6 sin(3x) cos(3x)−6 sin(3x) cos(3x)=0

Step 6: Therefore, the derivative of y= sin2(3x) + cos2(3x) with respect to

xis 0.

Question 27

Question

Find the derivative of the function f(x) = cos3(2x) + 2 sin2(x).

Solution

Step 1: Apply the chain rule to differentiate cos3(2x).

f′(x) = d

dx cos3(2x)+d

dx 2 sin2(x)

= 3 cos2(2x)·d

dx (cos(2x)) + 2 ·2 sin(x)·d

dx (sin(x))

Step 2: Differentiate cos(2x) and sin(x) using the chain rule.

f′(x) = 3 cos2(2x)·(−2 sin(2x)) + 4 sin(x) cos(x)

=−6 cos2(2x) sin(2x) + 4 sin(x) cos(x)

Step 3: Simplify the expression.

f′(x) = −6 cos2(2x) sin(2x) + 4 sin(x) cos(x)

Therefore, the derivative of the function f(x) = cos3(2x) + 2 sin2(x) is

f′(x) = −6 cos2(2x) sin(2x) + 4 sin(x) cos(x).

Question 28

Question

Find the derivative of the following function: f(x) = sin(x) cos(x).

Solution

Step 1: Recall the product rule, which states that the derivative of the product

of two functions u(x) and v(x) is given by:

(u·v)′=u′v+uv′.

In this case, let u(x) = sin(x) and v(x) = cos(x).

Step 2: Find u′(x) and v′(x):

u′(x) = cos(x)

v′(x) = −sin(x)

Step 3: Apply the product rule to find f′(x):

f′(x) = (sin(x)·cos(x))′= sin(x)·(−sin(x)) + cos(x)·cos(x)

f′(x) = −sin2(x) + cos2(x)

Step 4: Recall the Pythagorean identity sin2(x) + cos2(x) = 1:

f′(x) = −1 + cos2(x) = cos2(x)−1

Therefore, the derivative of f(x) = sin(x) cos(x) is f′(x) = cos2(x)−1.

Question 29

Question

Find the derivative of the function f(x) = sin2(x) cos3(x).

Solution

Step 1: Apply the product rule to differentiate the function f(x). Step 2: Recall

the product rule states that the derivative of the product of two functions is

the derivative of the first function times the second function plus the first func-

tion times the derivative of the second function. Step 3: Let u= sin2(x) and

v= cos3(x). Step 4: Find u′and v′. Step 5: u′= 2 sin(x) cos(x) by applying

the chain rule. Step 6: v′=−3 cos2(x) sin(x) by applying the chain rule. Step

7: Substitute u,u′,v, and v′into the product rule formula: f′(x) = u′v+uv′.

Step 8: Substituting all the values, we get f′(x) = (2 sin(x) cos(x))(cos3(x)) +

(sin2(x))(−3 cos2(x) sin(x)). Step 9: Simplify the expression. Step 10: Expand

the terms and simplify. Step 11: f′(x) = 2 sin(x) cos(x) cos3(x)−3 sin2(x) cos2(x) sin(x).

Step 12: Use trigonometric identities to simplify further (e.g., sin(2θ) = 2 sin(θ) cos(θ)).

Step 13: Simplify the expression to get the final answer for f′(x). Step 14: The

derivative of f(x) = sin2(x) cos3(x) is f′(x) = 2 sin(x) cos4(x)−3 sin2(x) cos2(x).

Question 30

Question

Find the derivative of the function f(x) = sin2(3x) cos(4x).

Solution

To find the derivative of f(x), we will use the product rule and the chain rule.

f(x) = sin2(3x) cos(4x)

Step 1: Apply the product rule. Let u= sin2(3x) and v= cos(4x).

f′(x) = u′v+uv′

Step 2: Find u′and v′.

u= sin2(3x)

u′= 2 sin(3x) cos(3x)·3

= 6 sin(3x) cos(3x)

v= cos(4x)

v′=−sin(4x)·4

=−4 sin(4x)

Step 3: Substitute u′,v′,u, and vinto the product rule formula and simplify.

f′(x) = (6 sin(3x) cos(3x)·cos(4x)) + (sin2(3x)· −4 sin(4x))

= 6 sin(3x) cos(3x) cos(4x)−4 sin2(3x) sin(4x)

Therefore, the derivative of f(x) = sin2(3x) cos(4x) is 6 sin(3x) cos(3x) cos(4x)−

4 sin2(3x) sin(4x).

Question 31

Question

Find the derivative of the function f(x) = sin(x)−tan(x)

cos(x).

Solution

Step 1: Rewrite the function using trigonometric identities. Step 2: Differentiate

each term in the function. Step 3: Simplify the derivatives obtained in Step 2.

Step 1:

Rewrite the function using trigonometric identities.

f(x) = sin(x)−tan(x)

cos(x)=sin(x)−sin(x)

cos(x)

cos(x)=sin(x) cos(x)−sin(x)

cos2(x)

Step 2:

Differentiate each term in the function.

f′(x) = d

dx sin(x) cos(x)−sin(x)

cos2(x)

=(cos(x) cos(x)−sin(x) sin(x)) cos2(x)−(sin(x) cos(x)−sin(x)(−sin(x) cos(x))

cos4(x)

Step 3:

Simplify the derivatives obtained in Step 2.

f′(x) = (cos2(x)−sin2(x)) cos2(x) + sin(x)2cos(x)

cos4(x)

=cos4(x)−sin2(x) cos2(x) + sin(x)2cos(x)

cos4(x)

= 1 −tan2(x) + sin(x) cos(x)

Therefore, the derivative of the function f(x) = sin(x)−tan(x)

cos(x)is f′(x) =

1−tan2(x) + sin(x) cos(x).

Question 32

Question

Find the derivative of f(x) = sin(x) cos(x).

Solution

Step 1: Recall the product rule for differentiation, which states that the deriva-

tive of the product of two functions is the derivative of the first function times

the second function plus the first function times the derivative of the second

function.

Step 2: Let’s apply the product rule to find the derivative of f(x) = sin(x) cos(x).

Step 3: Let u(x) = sin(x) and v(x) = cos(x).

Step 4: Using the product rule, we have:

f′(x) = u′(x)v(x) + u(x)v′(x)

Step 5: Now, find the derivatives of u(x) and v(x).

Step 6: u′(x) = cos(x) and v′(x) = −sin(x).

Step 7: Substitute these derivatives back into the product rule formula:

f′(x) = (cos(x) cos(x)) + (sin(x)(−sin(x)))

Step 8: Simplify the expression:

f′(x) = cos2(x)−sin2(x)

Step 9: Recall the Pythagorean identity: sin2(x) + cos2(x) = 1.

Step 10: Substitute the Pythagorean identity into the expression:

f′(x) = 1 −2 sin2(x)

Therefore, the derivative of f(x) = sin(x) cos(x) is f′(x)=1−2 sin2(x).

Question 33

Question

Calculate the derivative of the function f(x) = sin2x2−3x.

Solution

To find the derivative of the function f(x) = sin2x2−3x, we will use the

chain rule for differentiation.

Step 1: Let u= 2x2−3xbe the inner function and v= sin(u) be the outer

function. Then the function f(x) can be written as v(u) = sin(u).

Step 2: Calculate the derivative of the inner function uwith respect to x:

dx =d

dx (2x2−3x)=4x−3.

Step 3: Calculate the derivative of the outer function vwith respect to u:

du =d

du (sin(u)) = cos(u).

Step 4: Apply the chain rule (product of derivatives) to find the derivative

of f(x):

dx =dv

du ·du

dx = cos2x2−3x·(4x−3).

Therefore, the derivative of the function f(x) = sin2x2−3xis cos2x2−3x·

(4x−3).

Question 34

Question

Find the derivative of the function f(x) = sin2(x) cos3(x).

Solution

Step 1: To find the derivative of f(x) = sin2(x) cos3(x), we will use the product

rule.

Step 2: The product rule states that if f(x) = u(x)v(x), then f′(x) =

u′(x)v(x) + u(x)v′(x), where u(x) and v(x) are differentiable functions.

Step 3: Let u(x) = sin2(x) and v(x) = cos3(x).

Step 4: Calculate the derivatives of u(x) and v(x):

u′(x) = 2 sin(x) cos(x)

v′(x) = 3 cos2(x)(−sin(x)) = −3 sin(x) cos2(x)

Step 5: Now, apply the product rule to find f′(x):

f′(x) = u′(x)v(x) + u(x)v′(x)

= (2 sin(x) cos(x))(cos3(x)) + (sin2(x))(−3 sin(x) cos2(x))

= 2 sin(x) cos(x) cos3(x)−3 sin3(x) cos2(x)

Therefore, the derivative of f(x) = sin2(x) cos3(x) is f′(x) = 2 sin(x) cos(x) cos3(x)−

3 sin3(x) cos2(x).

Question 35

Question

Find the derivative of the function f(x) = sinx2+ cos(x).

Question 2

Question

Find the derivative of the function f(x) = cos2(3x) + sin(2x).

Solution

To find the derivative of f(x), we will use the chain rule and the sum rule for

differentiation.

Step 1: Find the derivative of cos2(3x) using the chain rule.

dx (cos2(3x)) = 2 cos(3x)·d

dx (cos(3x))

= 2 cos(3x)(−3 sin(3x))

=−6 cos(3x) sin(3x)

Step 2: Find the derivative of sin(2x).

dx (sin(2x)) = 2 cos(2x)

Step 3: Putting it all together, find the derivative of f(x).

f′(x) = d

dx (cos2(3x) + sin(2x))

=−6 cos(3x) sin(3x) + 2 cos(2x)

Question 3

Question

Find the derivative of the function f(x) = √3 cos(x)−sin(x)

cos(x).

Solution

Step 1: Let’s simplify the given function before finding the derivative. Step 2:

We can simplify f(x) by multiplying the numerator and denominator by cos(x).

Step 3: After simplifying, the function becomes f(x) = √3−tan(x). Step 4:

Now, we can find the derivative of f(x). Step 5: Recall that the derivative of a

constant is zero, and the derivative of tan(x) is sec2(x). Step 6: Therefore, the

derivative of f(x) is f′(x)=0−sec2(x). Step 7: Simplifying further, we get

f′(x) = −sec2(x).