MATH 332 - Quotient rule Question Bank
Question 1
Consider the function f(x) = ex
x2+1 . Use the quotient rule to find the deriva-
tive f′(x).
Solution:
Step 1: Identify the functions uand v- Let u(x) = ex- Let v(x) = x2+ 1
Step 2: Differentiate uand vwith respect to x-u′(x) = d
dx (ex) = ex-
v′(x) = d
dx (x2+ 1) = 2x
Step 3: Apply the Quotient Rule The Quotient Rule states that if f(x) =
u(x)
v(x), then:
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
Step 4: Substitute the derivatives and functions into the Quotient Rule
formula
f′(x) = ex(x2+ 1) −ex(2x)
(x2+ 1)2
=ex(x2+ 1 −2x)
(x2+ 1)2
Step 5: Simplify the expression
f′(x) = ex(x2−2x+ 1)
(x2+ 1)2
This final expression is the derivative of f(x) using the Quotient Rule. Ques-
tion 1: Quotient Rule Application
Consider the function f(x) = ex
x2+1 . Use the quotient rule to find
the derivative f′(x).
Solution:
Step 1: Identify the functions uand v- Let u(x) = ex- Let v(x) =
x2+ 1
Step 2: Differentiate uand vwith respect to x-u′(x) = d
dx (ex) = ex
-v′(x) = d
dx (x2+ 1) = 2x
Step 3: Apply the Quotient Rule The Quotient Rule states that if
f(x) = u(x)
v(x), then:
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
1
Step 4: Substitute the derivatives and functions into the Quotient
Rule formula
f′(x) = ex(x2+ 1) −ex(2x)
(x2+ 1)2
=ex(x2+ 1 −2x)
(x2+ 1)2
Step 5: Simplify the expression
f′(x) = ex(x2−2x+ 1)
(x2+ 1)2
This final expression is the derivative of f(x)using the Quotient
Rule.
Question 2
Calculate the derivative of the following function using the Quo-
tient Rule:
f(x) = x3+ 2x
x2−1
Step-by-Step Solution:
1. Identify the functions: Let’s denote the numerator as u(x)and
the denominator as v(x):
u(x) = x3+ 2x
v(x) = x2−1
2. Calculate the derivatives of u(x)and v(x):
u′(x) = d
dx (x3+ 2x)=3x2+ 2
v′(x) = d
dx (x2−1) = 2x
3. Apply the Quotient Rule: The derivative of a function in the
form of f(x) = u(x)
v(x)is given by:
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
Plugging in the derivatives and functions:
f′(x) = (3x2+ 2)(x2−1) −(x3+ 2x)(2x)
(x2−1)2
2
4. Simplify the expression: Expand the numerator:
(3x2+ 2)(x2−1) = 3x4−3x2+ 2x2−2=3x4−x2−2
(x3+ 2x)(2x) = 2x4+ 4x2
Subtract:
3x4−x2−2−2x4−4x2=x4−5x2−2
So, putting it all together:
f′(x) = x4−5x2−2
(x2−1)2
This is f′(x), the derivative of the given function using the quotient
rule. Question 2:
Calculate the derivative of the following function using the Quo-
tient Rule:
f(x) = x3+ 2x
x2−1
Step-by-Step Solution:
1. Identify the functions: Let’s denote the numerator as u(x)and
the denominator as v(x):
u(x) = x3+ 2x
v(x) = x2−1
2. Calculate the derivatives of u(x)and v(x):
u′(x) = d
dx (x3+ 2x)=3x2+ 2
v′(x) = d
dx (x2−1) = 2x
3. Apply the Quotient Rule: The derivative of a function in the
form of f(x) = u(x)
v(x)is given by:
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
Plugging in the derivatives and functions:
f′(x) = (3x2+ 2)(x2−1) −(x3+ 2x)(2x)
(x2−1)2
4. Simplify the expression: Expand the numerator:
(3x2+ 2)(x2−1) = 3x4−3x2+ 2x2−2=3x4−x2−2
3
(x3+ 2x)(2x)=2x4+ 4x2
Subtract:
3x4−x2−2−2x4−4x2=x4−5x2−2
So, putting it all together:
f′(x) = x4−5x2−2
(x2−1)2
This is f′(x), the derivative of the given function using the quotient
rule.
Question 3
Problem Statement: If f(x) = sin(x)
x2+1 , find f′(x)using the quotient
rule.
—
Solution
To find f′(x), where f(x) = u
vand u= sin(x)and v=x2+1, we utilize
the quotient rule. The quotient rule states:
f′(x) = u′v−uv′
v2
Step 1: Differentiate uand v-u= sin(x)-u′= cos(x)-v=x2+ 1 -
v′= 2x
Step 2: Apply the Quotient Rule - Substitute u′,u,v′, and vinto
the quotient rule formula:
f′(x) = cos(x)(x2+ 1) −sin(x)(2x)
(x2+ 1)2
Step 3: Simplify the expression - Expand and simplify the numer-
ator:
f′(x) = cos(x)x2+ cos(x)−2xsin(x)
x2+ 1)2
Final Answer: - The derivative of the function f(x) = sin(x)
x2+1 is:
f′(x) = x2cos(x) + cos(x)−2xsin(x)
(x2+ 1)2
This derivative will be valid for all x for which the function is
defined, except possibly at points where the denominator is zero
(though in this case, x2+ 1 never equals zero). Question 3: Quo-
tient Rule
Problem Statement: If f(x) = sin(x)
x2+1 , find f′(x)using the quotient
rule.
4
—
Solution
To find f′(x), where f(x) = u
vand u= sin(x)and v=x2+1, we utilize
the quotient rule. The quotient rule states:
f′(x) = u′v−uv′
v2
Step 1: Differentiate uand v-u= sin(x)-u′= cos(x)-v=x2+ 1 -
v′= 2x
Step 2: Apply the Quotient Rule - Substitute u′,u,v′, and vinto
the quotient rule formula:
f′(x) = cos(x)(x2+ 1) −sin(x)(2x)
(x2+ 1)2
Step 3: Simplify the expression - Expand and simplify the numer-
ator:
f′(x) = cos(x)x2+ cos(x)−2xsin(x)
x2+ 1)2
Final Answer: - The derivative of the function f(x) = sin(x)
x2+1 is:
f′(x) = x2cos(x) + cos(x)−2xsin(x)
(x2+ 1)2
This derivative will be valid for all x for which the function is
defined, except possibly at points where the denominator is zero
(though in this case, x2+ 1 never equals zero).
Question 4
Problem Statement: Calculate the derivative of the function f(x) =
x2+3x−4
x−2using the quotient rule.
—
Step-by-Step Solution:
Step 1: Identify the Numerator and Denominator
From the function f(x) = x2+3x−4
x−2: - Numerator u(x) = x2+ 3x−4-
Denominator v(x) = x−2
Step 2: Differentiate the Numerator and Denominator
Differentiate u(x):
u′(x) = d
dx (x2+ 3x−4) = 2x+ 3
Differentiate v(x):
v′(x) = d
dx (x−2) = 1
5
Step 3: Apply the Quotient Rule
The quotient rule states:
u
v′
=u′v−uv′
v2
Substituting in the derivatives:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the Expression
Expand the terms in the numerator:
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Now, subtract the second product from the first:
(2x2−x−6) −(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Thus, we have:
f′(x) = x2−4x−2
(x−2)2
Conclusion:
The derivative of the function f(x) = x2+3x−4
x−2using the quotient
rule is:
f′(x) = x2−4x−2
(x−2)2
This result provides the rate of change of the function at any
point x= 2, where the function is defined. Question 4: Quotient Rule
Application for Liberty University Calculus
Problem Statement: Calculate the derivative of the function f(x) =
x2+3x−4
x−2using the quotient rule.
—
Step-by-Step Solution:
Step 1: Identify the Numerator and Denominator
From the function f(x) = x2+3x−4
x−2: - Numerator u(x) = x2+ 3x−4-
Denominator v(x) = x−2
Step 2: Differentiate the Numerator and Denominator
Differentiate u(x):
u′(x) = d
dx (x2+ 3x−4) = 2x+ 3
Differentiate v(x):
v′(x) = d
dx (x−2) = 1
6
Step 3: Apply the Quotient Rule
The quotient rule states:
u
v′
=u′v−uv′
v2
Substituting in the derivatives:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the Expression
Expand the terms in the numerator:
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Now, subtract the second product from the first:
(2x2−x−6) −(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Thus, we have:
f′(x) = x2−4x−2
(x−2)2
Conclusion:
The derivative of the function f(x) = x2+3x−4
x−2using the quotient
rule is:
f′(x) = x2−4x−2
(x−2)2
This result provides the rate of change of the function at any point
x= 2, where the function is defined.
Question 5
Given the functions f(x) = e2xand g(x) = cos(x), find the derivative
of the quotient f(x)
g(x)using the Quotient Rule.
—
Solution:
Step 1: Identify f(x)and g(x)
You are given f(x) = e2xand g(x) = cos(x).
Step 2: Find f′(x)and g′(x)
Differentiate f(x):
f′(x) = d
dx [e2x]=2e2x
7
Differentiate g(x):
g′(x) = d
dx [cos(x)] = −sin(x)
Step 3: Apply the Quotient Rule
The Quotient Rule states:
f
g′
=f′g−fg′
g2
Substitute f(x),g(x),f′(x), and g′(x)into the formula:
e2x
cos(x)′
=(2e2x)(cos(x)) −(e2x)(−sin(x))
(cos(x))2
Step 4: Simplify the expression
Combine terms in the numerator:
=2e2xcos(x) + e2xsin(x)
cos2(x)
Extract common factor e2xin the numerator:
=e2x(2 cos(x) + sin(x))
cos2(x)
Step 5: Final answer
The derivative of the given function e2x
cos(x)is:
e2x
cos(x)′
=e2x(2 cos(x) + sin(x))
cos2(x)
This provides a step-by-step solution to finding the derivative using
the Quotient Rule. Question 5: Quotient Rule Application
Given the functions f(x) = e2xand g(x) = cos(x), find the derivative
of the quotient f(x)
g(x)using the Quotient Rule.
—
Solution:
Step 1: Identify f(x)and g(x)
You are given f(x) = e2xand g(x) = cos(x).
Step 2: Find f′(x)and g′(x)
Differentiate f(x):
f′(x) = d
dx [e2x]=2e2x
Differentiate g(x):
g′(x) = d
dx [cos(x)] = −sin(x)
8
Step 3: Apply the Quotient Rule
The Quotient Rule states:
f
g′
=f′g−fg′
g2
Substitute f(x),g(x),f′(x), and g′(x)into the formula:
e2x
cos(x)′
=(2e2x)(cos(x)) −(e2x)(−sin(x))
(cos(x))2
Step 4: Simplify the expression
Combine terms in the numerator:
=2e2xcos(x) + e2xsin(x)
cos2(x)
Extract common factor e2xin the numerator:
=e2x(2 cos(x) + sin(x))
cos2(x)
Step 5: Final answer
The derivative of the given function e2x
cos(x)is:
e2x
cos(x)′
=e2x(2 cos(x) + sin(x))
cos2(x)
This provides a step-by-step solution to finding the derivative using
the Quotient Rule.
Question 6
Given the function f(x) = x2+3x−4
x−2, use the Quotient Rule to find
f′(x).
Solution
Step 1: Identify the functions u(x)and v(x)To apply the Quotient
Rule, we need to identify the numerator and denominator functions.
Let u(x) = x2+ 3x−4and v(x) = x−2.
Step 2: Differentiate u(x)and v(x)The derivatives of u(x)and v(x)
are calculated as follows:
u′(x) = 2x+ 3
v′(x)=1
Step 3: Apply the Quotient Rule The Quotient Rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
9
Substitute the derivatives and functions into the formula:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the numerator Expand and simplify the expres-
sion in the numerator.
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Subtract the second expanded product from the first:
2x2−x−6−(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Step 5: Write the final derivative expression
f′(x) = x2−4x−2
(x−2)2
This is the derivative of the function f(x) = x2+3x−4
x−2using the
Quotient Rule. Question 6
Given the function f(x) = x2+3x−4
x−2, use the Quotient Rule to find
f′(x).
Solution
Step 1: Identify the functions u(x)and v(x)To apply the Quotient
Rule, we need to identify the numerator and denominator functions.
Let u(x) = x2+ 3x−4and v(x) = x−2.
Step 2: Differentiate u(x)and v(x)The derivatives of u(x)and v(x)
are calculated as follows:
u′(x) = 2x+ 3
v′(x)=1
Step 3: Apply the Quotient Rule The Quotient Rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
Substitute the derivatives and functions into the formula:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the numerator Expand and simplify the expres-
sion in the numerator.
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
10
(x2+ 3x−4)(1) = x2+ 3x−4
Subtract the second expanded product from the first:
2x2−x−6−(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Step 5: Write the final derivative expression
f′(x) = x2−4x−2
(x−2)2
This is the derivative of the function f(x) = x2+3x−4
x−2using the
Quotient Rule.
Question 7
Given the function f(x) = x3+3x2+5
x2, use the quotient rule to find
the derivative f′(x).
Solution:
Step 1: Identify the functions Let u(x) = x3+ 3x2+ 5 and v(x) = x2.
We need to find the derivative using the quotient rule.
Step 2: Apply the Quotient Rule The quotient rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
First, find u′(x)and v′(x):
-u′(x) = d
dx (x3+ 3x2+ 5) = 3x2+ 6x-v′(x) = d
dx (x2)=2x
Step 3: Plug into the formula Plugging these into the quotient
rule gives:
f′(x) = (3x2+ 6x)(x2)−(x3+ 3x2+ 5)(2x)
(x2)2
Simplify the expression inside the numerator:
f′(x) = 3x4+ 6x3−2x4−6x3−10x
x4
f′(x) = x4−10x
x4
Step 4: Simplify the derivative The derivative can be simplified
further:
f′(x) = x−3−10x−3
f′(x) = −9x−3
Step 5: Write in a more common form
f′(x) = −9
x3
11
This is the derivative of the function f(x)using the quotient rule.
Question 7: Quotient Rule Application
Given the function f(x) = x3+3x2+5
x2, use the quotient rule to find
the derivative f′(x).
Solution:
Step 1: Identify the functions Let u(x) = x3+ 3x2+ 5 and v(x) = x2.
We need to find the derivative using the quotient rule.
Step 2: Apply the Quotient Rule The quotient rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
First, find u′(x)and v′(x):
-u′(x) = d
dx (x3+ 3x2+ 5) = 3x2+ 6x-v′(x) = d
dx (x2)=2x
Step 3: Plug into the formula Plugging these into the quotient
rule gives:
f′(x) = (3x2+ 6x)(x2)−(x3+ 3x2+ 5)(2x)
(x2)2
Simplify the expression inside the numerator:
f′(x) = 3x4+ 6x3−2x4−6x3−10x
x4
f′(x) = x4−10x
x4
Step 4: Simplify the derivative The derivative can be simplified
further:
f′(x) = x−3−10x−3
f′(x) = −9x−3
Step 5: Write in a more common form
f′(x) = −9
x3
This is the derivative of the function f(x)using the quotient rule.
Question 8
Calculate the derivative of the function f(x) = x3+2x
x2−5.
Solution:
To find the derivative of the given function, we use the Quotient
Rule. The Quotient Rule states that if you have a function defined
as f(x) = g(x)
h(x), then its derivative f′(x)is given by:
f′(x) = g′(x)h(x)−g(x)h′(x)
[h(x)]2
12
Step 1: Identify g(x)and h(x).
In the function f(x) = x3+2x
x2−5: - g(x) = x3+ 2x-h(x) = x2−5
Step 2: Differentiate g(x)and h(x).
- Derivative of g(x) = x3+ 2x:
g′(x)=3x2+ 2
- Derivative of h(x) = x2−5:
h′(x)=2x
Step 3: Apply the Quotient Rule.
Plug in g(x),g′(x),h(x), and h′(x)into the Quotient Rule formula:
f′(x) = (3x2+ 2)(x2−5) −(x3+ 2x)(2x)
(x2−5)2
f′(x) = 3x4−15x2+ 2x2−10 −2x4−4x
(x2−5)2
f′(x) = x4−17x2−4x−10
(x2−5)2
Step 4: Simplify the expression (if possible).
In this case, the expression for f′(x)is already in a relatively sim-
plified form:
f′(x) = x4−17x2−4x−10
(x2−5)2
By following these steps using the Quotient Rule, we have success-
fully calculated the derivative for the function f(x) = x3+2x
x2−5. Question
8: Applying the Quotient Rule
Calculate the derivative of the function f(x) = x3+2x
x2−5.
Solution:
To find the derivative of the given function, we use the Quotient
Rule. The Quotient Rule states that if you have a function defined
as f(x) = g(x)
h(x), then its derivative f′(x)is given by:
f′(x) = g′(x)h(x)−g(x)h′(x)
[h(x)]2
Step 1: Identify g(x)and h(x).
In the function f(x) = x3+2x
x2−5: - g(x) = x3+ 2x-h(x) = x2−5
Step 2: Differentiate g(x)and h(x).
- Derivative of g(x) = x3+ 2x:
g′(x)=3x2+ 2
13
- Derivative of h(x) = x2−5:
h′(x)=2x
Step 3: Apply the Quotient Rule.
Plug in g(x),g′(x),h(x), and h′(x)into the Quotient Rule formula:
f′(x) = (3x2+ 2)(x2−5) −(x3+ 2x)(2x)
(x2−5)2
f′(x) = 3x4−15x2+ 2x2−10 −2x4−4x
(x2−5)2
f′(x) = x4−17x2−4x−10
(x2−5)2
Step 4: Simplify the expression (if possible).
In this case, the expression for f′(x)is already in a relatively sim-
plified form:
f′(x) = x4−17x2−4x−10
(x2−5)2
By following these steps using the Quotient Rule, we have success-
fully calculated the derivative for the function f(x) = x3+2x
x2−5.
Question 9
Problem:
Given the functions f(x) = x3−4x2+ 6x−2and g(x) = x−2, use the
quotient rule to find the derivative of the function h(x) = f(x)
g(x).
—
Solution:
Step 1: Identify the functions involved in the quotient rule
We have: - f(x) = x3−4x2+ 6x−2-g(x) = x−2
Step 2: Write down the formula for the quotient rule
The quotient rule states that if h(x) = f(x)
g(x), then the derivative
h′(x)is given by:
h′(x) = f′(x)g(x)−f(x)g′(x)
[g(x)]2
Step 3: Find the derivatives f′(x)and g′(x)
- Derivative of f(x) = x3−4x2+ 6x−2:
f′(x)=3x2−8x+ 6
- Derivative of g(x) = x−2:
g′(x) = 1
14
Step 4: Substitute f(x),g(x),f′(x), and g′(x)into the quotient rule
formula
h′(x) = (3x2−8x+ 6)(x−2) −(x3−4x2+ 6x−2)(1)
(x−2)2
Step 5: Simplify the numerator
Multiply out the terms in the numerator:
(3x2−8x+ 6)(x−2) = 3x3−6x2−8x2+ 16x+ 6x−12
= 3x3−14x2+ 22x−12
Subtract f(x)multiplied by g′(x):
3x3−14x2+ 22x−12 −(x3−4x2+ 6x−2)
= 3x3−14x2+ 22x−12 −x3+ 4x2−6x+ 2
= 2x3−10x2+ 16x−10
Step 6: Write down the final expression for h′(x)
h′(x) = 2x3−10x2+ 16x−10
(x−2)2
Thus, the derivative of the function h(x) = f(x)
g(x)is h′(x) = 2x3−10x2+16x−10
(x−2)2.
Question 9: Quotient Rule Application
Problem:
Given the functions f(x) = x3−4x2+ 6x−2and g(x) = x−2, use the
quotient rule to find the derivative of the function h(x) = f(x)
g(x).
—
Solution:
Step 1: Identify the functions involved in the quotient rule
We have: - f(x) = x3−4x2+ 6x−2-g(x) = x−2
Step 2: Write down the formula for the quotient rule
The quotient rule states that if h(x) = f(x)
g(x), then the derivative
h′(x)is given by:
h′(x) = f′(x)g(x)−f(x)g′(x)
[g(x)]2
Step 3: Find the derivatives f′(x)and g′(x)
- Derivative of f(x) = x3−4x2+ 6x−2:
f′(x)=3x2−8x+ 6
- Derivative of g(x) = x−2:
g′(x) = 1
15
Step 4: Substitute f(x),g(x),f′(x), and g′(x)into the quotient rule
formula
h′(x) = (3x2−8x+ 6)(x−2) −(x3−4x2+ 6x−2)(1)
(x−2)2
Step 5: Simplify the numerator
Multiply out the terms in the numerator:
(3x2−8x+ 6)(x−2) = 3x3−6x2−8x2+ 16x+ 6x−12
= 3x3−14x2+ 22x−12
Subtract f(x)multiplied by g′(x):
3x3−14x2+ 22x−12 −(x3−4x2+ 6x−2)
= 3x3−14x2+ 22x−12 −x3+ 4x2−6x+ 2
= 2x3−10x2+ 16x−10
Step 6: Write down the final expression for h′(x)
h′(x) = 2x3−10x2+ 16x−10
(x−2)2
Thus, the derivative of the function h(x) = f(x)
g(x)is h′(x) = 2x3−10x2+16x−10
(x−2)2.
Question 10
Problem: Given the functions f(x) = x2+ 5x+ 3 and g(x)=2x+ 1,
find the derivative d
dx f(x)
g(x).
Solution, Step-by-Step:
Step 1: Identify the functions f(x)and g(x). - f(x) = x2+ 5x+ 3 -
g(x)=2x+ 1
Step 2: Compute the derivatives f′(x)and g′(via). - For f(x):
f′(x) = d
dx (x2+ 5x+ 3) = 2x+ 5
- For g(x):
g′(x) = d
dx (2x+ 1) = 2
Step 3: Apply the quotient rule. The quotient rule states that
u
v′=u′v−uv′
v2, where u=f(x)and v=g(x).
- Plug f(x),g(x),f′(x), and g′(x)into the formula:
f(x)
g(x)′
=(2x+ 5)(2x+ 1) −(x2+ 5x+ 3)(2)
(2x+ 1)2
16
Step 4: Simplify the numerator. - Expand and simplify:
(2x+ 5)(2x+ 1) = 4x2+ 2x+ 10x+ 5 = 4x2+ 12x+ 5
(x2+ 5x+ 3)(2) = 2x2+ 10x+ 6
4x2+ 12x+ 5 −(2x2+ 10x+ 6) = 4x2+ 12x+ 5 −2x2−10x−6
= 2x2+ 2x−1
Step 5: Write the final derivative expression. - Put the simplified
expression over the squared denominator:
f(x)
g(x)′
=2x2+ 2x−1
(2x+ 1)2
This is the derivative of the function f(x)
g(x)using the quotient rule.
Question 10: Quotient Rule
Problem: Given the functions f(x) = x2+ 5x+ 3 and g(x)=2x+ 1,
find the derivative d
dx f(x)
g(x).
Solution, Step-by-Step:
Step 1: Identify the functions f(x)and g(x). - f(x) = x2+ 5x+ 3 -
g(x)=2x+ 1
Step 2: Compute the derivatives f′(x)and g′(via). - For f(x):
f′(x) = d
dx (x2+ 5x+ 3) = 2x+ 5
- For g(x):
g′(x) = d
dx (2x+ 1) = 2
Step 3: Apply the quotient rule. The quotient rule states that
u
v′=u′v−uv′
v2, where u=f(x)and v=g(x).
- Plug f(x),g(x),f′(x), and g′(x)into the formula:
f(x)
g(x)′
=(2x+ 5)(2x+ 1) −(x2+ 5x+ 3)(2)
(2x+ 1)2
Step 4: Simplify the numerator. - Expand and simplify:
(2x+ 5)(2x+ 1) = 4x2+ 2x+ 10x+ 5 = 4x2+ 12x+ 5
(x2+ 5x+ 3)(2) = 2x2+ 10x+ 6
4x2+ 12x+ 5 −(2x2+ 10x+ 6) = 4x2+ 12x+ 5 −2x2−10x−6
= 2x2+ 2x−1
Step 5: Write the final derivative expression. - Put the simplified
expression over the squared denominator:
f(x)
g(x)′
=2x2+ 2x−1
(2x+ 1)2
This is the derivative of the function f(x)
g(x)using the quotient rule.
17
(x3+ 2x)(2x)=2x4+ 4x2
Subtract:
3x4−x2−2−2x4−4x2=x4−5x2−2
So, putting it all together:
f′(x) = x4−5x2−2
(x2−1)2
This is f′(x), the derivative of the given function using the quotient
rule.
Question 3
Problem Statement: If f(x) = sin(x)
x2+1 , find f′(x)using the quotient
rule.
—
Solution
To find f′(x), where f(x) = u
vand u= sin(x)and v=x2+1, we utilize
the quotient rule. The quotient rule states:
f′(x) = u′v−uv′
v2
Step 1: Differentiate uand v-u= sin(x)-u′= cos(x)-v=x2+ 1 -
v′= 2x
Step 2: Apply the Quotient Rule - Substitute u′,u,v′, and vinto
the quotient rule formula:
f′(x) = cos(x)(x2+ 1) −sin(x)(2x)
(x2+ 1)2
Step 3: Simplify the expression - Expand and simplify the numer-
ator:
f′(x) = cos(x)x2+ cos(x)−2xsin(x)
x2+ 1)2
Final Answer: - The derivative of the function f(x) = sin(x)
x2+1 is:
f′(x) = x2cos(x) + cos(x)−2xsin(x)
(x2+ 1)2
This derivative will be valid for all x for which the function is
defined, except possibly at points where the denominator is zero
(though in this case, x2+ 1 never equals zero). Question 3: Quo-
tient Rule
Problem Statement: If f(x) = sin(x)
x2+1 , find f′(x)using the quotient
rule.
4
—
Solution
To find f′(x), where f(x) = u
vand u= sin(x)and v=x2+1, we utilize
the quotient rule. The quotient rule states:
f′(x) = u′v−uv′
v2
Step 1: Differentiate uand v-u= sin(x)-u′= cos(x)-v=x2+ 1 -
v′= 2x
Step 2: Apply the Quotient Rule - Substitute u′,u,v′, and vinto
the quotient rule formula:
f′(x) = cos(x)(x2+ 1) −sin(x)(2x)
(x2+ 1)2
Step 3: Simplify the expression - Expand and simplify the numer-
ator:
f′(x) = cos(x)x2+ cos(x)−2xsin(x)
x2+ 1)2
Final Answer: - The derivative of the function f(x) = sin(x)
x2+1 is:
f′(x) = x2cos(x) + cos(x)−2xsin(x)
(x2+ 1)2
This derivative will be valid for all x for which the function is
defined, except possibly at points where the denominator is zero
(though in this case, x2+ 1 never equals zero).
Question 4
Problem Statement: Calculate the derivative of the function f(x) =
x2+3x−4
x−2using the quotient rule.
—
Step-by-Step Solution:
Step 1: Identify the Numerator and Denominator
From the function f(x) = x2+3x−4
x−2: - Numerator u(x) = x2+ 3x−4-
Denominator v(x) = x−2
Step 2: Differentiate the Numerator and Denominator
Differentiate u(x):
u′(x) = d
dx (x2+ 3x−4) = 2x+ 3
Differentiate v(x):
v′(x) = d
dx (x−2) = 1
5
Step 3: Apply the Quotient Rule
The quotient rule states:
u
v′
=u′v−uv′
v2
Substituting in the derivatives:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the Expression
Expand the terms in the numerator:
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Now, subtract the second product from the first:
(2x2−x−6) −(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Thus, we have:
f′(x) = x2−4x−2
(x−2)2
Conclusion:
The derivative of the function f(x) = x2+3x−4
x−2using the quotient
rule is:
f′(x) = x2−4x−2
(x−2)2
This result provides the rate of change of the function at any
point x= 2, where the function is defined. Question 4: Quotient Rule
Application for Liberty University Calculus
Problem Statement: Calculate the derivative of the function f(x) =
x2+3x−4
x−2using the quotient rule.
—
Step-by-Step Solution:
Step 1: Identify the Numerator and Denominator
From the function f(x) = x2+3x−4
x−2: - Numerator u(x) = x2+ 3x−4-
Denominator v(x) = x−2
Step 2: Differentiate the Numerator and Denominator
Differentiate u(x):
u′(x) = d
dx (x2+ 3x−4) = 2x+ 3
Differentiate v(x):
v′(x) = d
dx (x−2) = 1
6
Step 3: Apply the Quotient Rule
The quotient rule states:
u
v′
=u′v−uv′
v2
Substituting in the derivatives:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the Expression
Expand the terms in the numerator:
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Now, subtract the second product from the first:
(2x2−x−6) −(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Thus, we have:
f′(x) = x2−4x−2
(x−2)2
Conclusion:
The derivative of the function f(x) = x2+3x−4
x−2using the quotient
rule is:
f′(x) = x2−4x−2
(x−2)2
This result provides the rate of change of the function at any point
x= 2, where the function is defined.
Question 5
Given the functions f(x) = e2xand g(x) = cos(x), find the derivative
of the quotient f(x)
g(x)using the Quotient Rule.
—
Solution:
Step 1: Identify f(x)and g(x)
You are given f(x) = e2xand g(x) = cos(x).
Step 2: Find f′(x)and g′(x)
Differentiate f(x):
f′(x) = d
dx [e2x]=2e2x
7
Differentiate g(x):
g′(x) = d
dx [cos(x)] = −sin(x)
Step 3: Apply the Quotient Rule
The Quotient Rule states:
f
g′
=f′g−fg′
g2
Substitute f(x),g(x),f′(x), and g′(x)into the formula:
e2x
cos(x)′
=(2e2x)(cos(x)) −(e2x)(−sin(x))
(cos(x))2
Step 4: Simplify the expression
Combine terms in the numerator:
=2e2xcos(x) + e2xsin(x)
cos2(x)
Extract common factor e2xin the numerator:
=e2x(2 cos(x) + sin(x))
cos2(x)
Step 5: Final answer
The derivative of the given function e2x
cos(x)is:
e2x
cos(x)′
=e2x(2 cos(x) + sin(x))
cos2(x)
This provides a step-by-step solution to finding the derivative using
the Quotient Rule. Question 5: Quotient Rule Application
Given the functions f(x) = e2xand g(x) = cos(x), find the derivative
of the quotient f(x)
g(x)using the Quotient Rule.
—
Solution:
Step 1: Identify f(x)and g(x)
You are given f(x) = e2xand g(x) = cos(x).
Step 2: Find f′(x)and g′(x)
Differentiate f(x):
f′(x) = d
dx [e2x]=2e2x
Differentiate g(x):
g′(x) = d
dx [cos(x)] = −sin(x)
8
Step 3: Apply the Quotient Rule
The Quotient Rule states:
f
g′
=f′g−fg′
g2
Substitute f(x),g(x),f′(x), and g′(x)into the formula:
e2x
cos(x)′
=(2e2x)(cos(x)) −(e2x)(−sin(x))
(cos(x))2
Step 4: Simplify the expression
Combine terms in the numerator:
=2e2xcos(x) + e2xsin(x)
cos2(x)
Extract common factor e2xin the numerator:
=e2x(2 cos(x) + sin(x))
cos2(x)
Step 5: Final answer
The derivative of the given function e2x
cos(x)is:
e2x
cos(x)′
=e2x(2 cos(x) + sin(x))
cos2(x)
This provides a step-by-step solution to finding the derivative using
the Quotient Rule.
Question 6
Given the function f(x) = x2+3x−4
x−2, use the Quotient Rule to find
f′(x).
Solution
Step 1: Identify the functions u(x)and v(x)To apply the Quotient
Rule, we need to identify the numerator and denominator functions.
Let u(x) = x2+ 3x−4and v(x) = x−2.
Step 2: Differentiate u(x)and v(x)The derivatives of u(x)and v(x)
are calculated as follows:
u′(x) = 2x+ 3
v′(x)=1
Step 3: Apply the Quotient Rule The Quotient Rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
9
Substitute the derivatives and functions into the formula:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the numerator Expand and simplify the expres-
sion in the numerator.
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Subtract the second expanded product from the first:
2x2−x−6−(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Step 5: Write the final derivative expression
f′(x) = x2−4x−2
(x−2)2
This is the derivative of the function f(x) = x2+3x−4
x−2using the
Quotient Rule. Question 6
Given the function f(x) = x2+3x−4
x−2, use the Quotient Rule to find
f′(x).
Solution
Step 1: Identify the functions u(x)and v(x)To apply the Quotient
Rule, we need to identify the numerator and denominator functions.
Let u(x) = x2+ 3x−4and v(x) = x−2.
Step 2: Differentiate u(x)and v(x)The derivatives of u(x)and v(x)
are calculated as follows:
u′(x) = 2x+ 3
v′(x)=1
Step 3: Apply the Quotient Rule The Quotient Rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
Substitute the derivatives and functions into the formula:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the numerator Expand and simplify the expres-
sion in the numerator.
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
10
(x2+ 3x−4)(1) = x2+ 3x−4
Subtract the second expanded product from the first:
2x2−x−6−(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Step 5: Write the final derivative expression
f′(x) = x2−4x−2
(x−2)2
This is the derivative of the function f(x) = x2+3x−4
x−2using the
Quotient Rule.
Question 7
Given the function f(x) = x3+3x2+5
x2, use the quotient rule to find
the derivative f′(x).
Solution:
Step 1: Identify the functions Let u(x) = x3+ 3x2+ 5 and v(x) = x2.
We need to find the derivative using the quotient rule.
Step 2: Apply the Quotient Rule The quotient rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
First, find u′(x)and v′(x):
-u′(x) = d
dx (x3+ 3x2+ 5) = 3x2+ 6x-v′(x) = d
dx (x2)=2x
Step 3: Plug into the formula Plugging these into the quotient
rule gives:
f′(x) = (3x2+ 6x)(x2)−(x3+ 3x2+ 5)(2x)
(x2)2
Simplify the expression inside the numerator:
f′(x) = 3x4+ 6x3−2x4−6x3−10x
x4
f′(x) = x4−10x
x4
Step 4: Simplify the derivative The derivative can be simplified
further:
f′(x) = x−3−10x−3
f′(x) = −9x−3
Step 5: Write in a more common form
f′(x) = −9
x3
11
This is the derivative of the function f(x)using the quotient rule.
Question 7: Quotient Rule Application
Given the function f(x) = x3+3x2+5
x2, use the quotient rule to find
the derivative f′(x).
Solution:
Step 1: Identify the functions Let u(x) = x3+ 3x2+ 5 and v(x) = x2.
We need to find the derivative using the quotient rule.
Step 2: Apply the Quotient Rule The quotient rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
First, find u′(x)and v′(x):
-u′(x) = d
dx (x3+ 3x2+ 5) = 3x2+ 6x-v′(x) = d
dx (x2)=2x
Step 3: Plug into the formula Plugging these into the quotient
rule gives:
f′(x) = (3x2+ 6x)(x2)−(x3+ 3x2+ 5)(2x)
(x2)2
Simplify the expression inside the numerator:
f′(x) = 3x4+ 6x3−2x4−6x3−10x
x4
f′(x) = x4−10x
x4
Step 4: Simplify the derivative The derivative can be simplified
further:
f′(x) = x−3−10x−3
f′(x) = −9x−3
Step 5: Write in a more common form
f′(x) = −9
x3
This is the derivative of the function f(x)using the quotient rule.
Question 8
Calculate the derivative of the function f(x) = x3+2x
x2−5.
Solution:
To find the derivative of the given function, we use the Quotient
Rule. The Quotient Rule states that if you have a function defined
as f(x) = g(x)
h(x), then its derivative f′(x)is given by:
f′(x) = g′(x)h(x)−g(x)h′(x)
[h(x)]2
12
Step 1: Identify g(x)and h(x).
In the function f(x) = x3+2x
x2−5: - g(x) = x3+ 2x-h(x) = x2−5
Step 2: Differentiate g(x)and h(x).
- Derivative of g(x) = x3+ 2x:
g′(x)=3x2+ 2
- Derivative of h(x) = x2−5:
h′(x)=2x
Step 3: Apply the Quotient Rule.
Plug in g(x),g′(x),h(x), and h′(x)into the Quotient Rule formula:
f′(x) = (3x2+ 2)(x2−5) −(x3+ 2x)(2x)
(x2−5)2
f′(x) = 3x4−15x2+ 2x2−10 −2x4−4x
(x2−5)2
f′(x) = x4−17x2−4x−10
(x2−5)2
Step 4: Simplify the expression (if possible).
In this case, the expression for f′(x)is already in a relatively sim-
plified form:
f′(x) = x4−17x2−4x−10
(x2−5)2
By following these steps using the Quotient Rule, we have success-
fully calculated the derivative for the function f(x) = x3+2x
x2−5. Question
8: Applying the Quotient Rule
Calculate the derivative of the function f(x) = x3+2x
x2−5.
Solution:
To find the derivative of the given function, we use the Quotient
Rule. The Quotient Rule states that if you have a function defined
as f(x) = g(x)
h(x), then its derivative f′(x)is given by:
f′(x) = g′(x)h(x)−g(x)h′(x)
[h(x)]2
Step 1: Identify g(x)and h(x).
In the function f(x) = x3+2x
x2−5: - g(x) = x3+ 2x-h(x) = x2−5
Step 2: Differentiate g(x)and h(x).
- Derivative of g(x) = x3+ 2x:
g′(x)=3x2+ 2
13
- Derivative of h(x) = x2−5:
h′(x)=2x
Step 3: Apply the Quotient Rule.
Plug in g(x),g′(x),h(x), and h′(x)into the Quotient Rule formula:
f′(x) = (3x2+ 2)(x2−5) −(x3+ 2x)(2x)
(x2−5)2
f′(x) = 3x4−15x2+ 2x2−10 −2x4−4x
(x2−5)2
f′(x) = x4−17x2−4x−10
(x2−5)2
Step 4: Simplify the expression (if possible).
In this case, the expression for f′(x)is already in a relatively sim-
plified form:
f′(x) = x4−17x2−4x−10
(x2−5)2
By following these steps using the Quotient Rule, we have success-
fully calculated the derivative for the function f(x) = x3+2x
x2−5.
Question 9
Problem:
Given the functions f(x) = x3−4x2+ 6x−2and g(x) = x−2, use the
quotient rule to find the derivative of the function h(x) = f(x)
g(x).
—
Solution:
Step 1: Identify the functions involved in the quotient rule
We have: - f(x) = x3−4x2+ 6x−2-g(x) = x−2
Step 2: Write down the formula for the quotient rule
The quotient rule states that if h(x) = f(x)
g(x), then the derivative
h′(x)is given by:
h′(x) = f′(x)g(x)−f(x)g′(x)
[g(x)]2
Step 3: Find the derivatives f′(x)and g′(x)
- Derivative of f(x) = x3−4x2+ 6x−2:
f′(x)=3x2−8x+ 6
- Derivative of g(x) = x−2:
g′(x) = 1
14
Step 4: Substitute f(x),g(x),f′(x), and g′(x)into the quotient rule
formula
h′(x) = (3x2−8x+ 6)(x−2) −(x3−4x2+ 6x−2)(1)
(x−2)2
Step 5: Simplify the numerator
Multiply out the terms in the numerator:
(3x2−8x+ 6)(x−2) = 3x3−6x2−8x2+ 16x+ 6x−12
= 3x3−14x2+ 22x−12
Subtract f(x)multiplied by g′(x):
3x3−14x2+ 22x−12 −(x3−4x2+ 6x−2)
= 3x3−14x2+ 22x−12 −x3+ 4x2−6x+ 2
= 2x3−10x2+ 16x−10
Step 6: Write down the final expression for h′(x)
h′(x) = 2x3−10x2+ 16x−10
(x−2)2
Thus, the derivative of the function h(x) = f(x)
g(x)is h′(x) = 2x3−10x2+16x−10
(x−2)2.
Question 9: Quotient Rule Application
Problem:
Given the functions f(x) = x3−4x2+ 6x−2and g(x) = x−2, use the
quotient rule to find the derivative of the function h(x) = f(x)
g(x).
—
Solution:
Step 1: Identify the functions involved in the quotient rule
We have: - f(x) = x3−4x2+ 6x−2-g(x) = x−2
Step 2: Write down the formula for the quotient rule
The quotient rule states that if h(x) = f(x)
g(x), then the derivative
h′(x)is given by:
h′(x) = f′(x)g(x)−f(x)g′(x)
[g(x)]2
Step 3: Find the derivatives f′(x)and g′(x)
- Derivative of f(x) = x3−4x2+ 6x−2:
f′(x)=3x2−8x+ 6
- Derivative of g(x) = x−2:
g′(x) = 1
15
Step 4: Substitute f(x),g(x),f′(x), and g′(x)into the quotient rule
formula
h′(x) = (3x2−8x+ 6)(x−2) −(x3−4x2+ 6x−2)(1)
(x−2)2
Step 5: Simplify the numerator
Multiply out the terms in the numerator:
(3x2−8x+ 6)(x−2) = 3x3−6x2−8x2+ 16x+ 6x−12
= 3x3−14x2+ 22x−12
Subtract f(x)multiplied by g′(x):
3x3−14x2+ 22x−12 −(x3−4x2+ 6x−2)
= 3x3−14x2+ 22x−12 −x3+ 4x2−6x+ 2
= 2x3−10x2+ 16x−10
Step 6: Write down the final expression for h′(x)
h′(x) = 2x3−10x2+ 16x−10
(x−2)2
Thus, the derivative of the function h(x) = f(x)
g(x)is h′(x) = 2x3−10x2+16x−10
(x−2)2.
Question 10
Problem: Given the functions f(x) = x2+ 5x+ 3 and g(x)=2x+ 1,
find the derivative d
dx f(x)
g(x).
Solution, Step-by-Step:
Step 1: Identify the functions f(x)and g(x). - f(x) = x2+ 5x+ 3 -
g(x)=2x+ 1
Step 2: Compute the derivatives f′(x)and g′(via). - For f(x):
f′(x) = d
dx (x2+ 5x+ 3) = 2x+ 5
- For g(x):
g′(x) = d
dx (2x+ 1) = 2
Step 3: Apply the quotient rule. The quotient rule states that
u
v′=u′v−uv′
v2, where u=f(x)and v=g(x).
- Plug f(x),g(x),f′(x), and g′(x)into the formula:
f(x)
g(x)′
=(2x+ 5)(2x+ 1) −(x2+ 5x+ 3)(2)
(2x+ 1)2
16
Step 4: Simplify the numerator. - Expand and simplify:
(2x+ 5)(2x+ 1) = 4x2+ 2x+ 10x+ 5 = 4x2+ 12x+ 5
(x2+ 5x+ 3)(2) = 2x2+ 10x+ 6
4x2+ 12x+ 5 −(2x2+ 10x+ 6) = 4x2+ 12x+ 5 −2x2−10x−6
= 2x2+ 2x−1
Step 5: Write the final derivative expression. - Put the simplified
expression over the squared denominator:
f(x)
g(x)′
=2x2+ 2x−1
(2x+ 1)2
This is the derivative of the function f(x)
g(x)using the quotient rule.
Question 10: Quotient Rule
Problem: Given the functions f(x) = x2+ 5x+ 3 and g(x)=2x+ 1,
find the derivative d
dx f(x)
g(x).
Solution, Step-by-Step:
Step 1: Identify the functions f(x)and g(x). - f(x) = x2+ 5x+ 3 -
g(x)=2x+ 1
Step 2: Compute the derivatives f′(x)and g′(via). - For f(x):
f′(x) = d
dx (x2+ 5x+ 3) = 2x+ 5
- For g(x):
g′(x) = d
dx (2x+ 1) = 2
Step 3: Apply the quotient rule. The quotient rule states that
u
v′=u′v−uv′
v2, where u=f(x)and v=g(x).
- Plug f(x),g(x),f′(x), and g′(x)into the formula:
f(x)
g(x)′
=(2x+ 5)(2x+ 1) −(x2+ 5x+ 3)(2)
(2x+ 1)2
Step 4: Simplify the numerator. - Expand and simplify:
(2x+ 5)(2x+ 1) = 4x2+ 2x+ 10x+ 5 = 4x2+ 12x+ 5
(x2+ 5x+ 3)(2) = 2x2+ 10x+ 6
4x2+ 12x+ 5 −(2x2+ 10x+ 6) = 4x2+ 12x+ 5 −2x2−10x−6
= 2x2+ 2x−1
Step 5: Write the final derivative expression. - Put the simplified
expression over the squared denominator:
f(x)
g(x)′
=2x2+ 2x−1
(2x+ 1)2
This is the derivative of the function f(x)
g(x)using the quotient rule.
17
(x3+ 2x)(2x)=2x4+ 4x2
Subtract:
3x4−x2−2−2x4−4x2=x4−5x2−2
So, putting it all together:
f′(x) = x4−5x2−2
(x2−1)2
This is f′(x), the derivative of the given function using the quotient
rule.
Question 3
Problem Statement: If f(x) = sin(x)
x2+1 , find f′(x)using the quotient
rule.
—
Solution
To find f′(x), where f(x) = u
vand u= sin(x)and v=x2+1, we utilize
the quotient rule. The quotient rule states:
f′(x) = u′v−uv′
v2
Step 1: Differentiate uand v-u= sin(x)-u′= cos(x)-v=x2+ 1 -
v′= 2x
Step 2: Apply the Quotient Rule - Substitute u′,u,v′, and vinto
the quotient rule formula:
f′(x) = cos(x)(x2+ 1) −sin(x)(2x)
(x2+ 1)2
Step 3: Simplify the expression - Expand and simplify the numer-
ator:
f′(x) = cos(x)x2+ cos(x)−2xsin(x)
x2+ 1)2
Final Answer: - The derivative of the function f(x) = sin(x)
x2+1 is:
f′(x) = x2cos(x) + cos(x)−2xsin(x)
(x2+ 1)2
This derivative will be valid for all x for which the function is
defined, except possibly at points where the denominator is zero
(though in this case, x2+ 1 never equals zero). Question 3: Quo-
tient Rule
Problem Statement: If f(x) = sin(x)
x2+1 , find f′(x)using the quotient
rule.
4
—
Solution
To find f′(x), where f(x) = u
vand u= sin(x)and v=x2+1, we utilize
the quotient rule. The quotient rule states:
f′(x) = u′v−uv′
v2
Step 1: Differentiate uand v-u= sin(x)-u′= cos(x)-v=x2+ 1 -
v′= 2x
Step 2: Apply the Quotient Rule - Substitute u′,u,v′, and vinto
the quotient rule formula:
f′(x) = cos(x)(x2+ 1) −sin(x)(2x)
(x2+ 1)2
Step 3: Simplify the expression - Expand and simplify the numer-
ator:
f′(x) = cos(x)x2+ cos(x)−2xsin(x)
x2+ 1)2
Final Answer: - The derivative of the function f(x) = sin(x)
x2+1 is:
f′(x) = x2cos(x) + cos(x)−2xsin(x)
(x2+ 1)2
This derivative will be valid for all x for which the function is
defined, except possibly at points where the denominator is zero
(though in this case, x2+ 1 never equals zero).
Question 4
Problem Statement: Calculate the derivative of the function f(x) =
x2+3x−4
x−2using the quotient rule.
—
Step-by-Step Solution:
Step 1: Identify the Numerator and Denominator
From the function f(x) = x2+3x−4
x−2: - Numerator u(x) = x2+ 3x−4-
Denominator v(x) = x−2
Step 2: Differentiate the Numerator and Denominator
Differentiate u(x):
u′(x) = d
dx (x2+ 3x−4) = 2x+ 3
Differentiate v(x):
v′(x) = d
dx (x−2) = 1
5
Step 3: Apply the Quotient Rule
The quotient rule states:
u
v′
=u′v−uv′
v2
Substituting in the derivatives:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the Expression
Expand the terms in the numerator:
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Now, subtract the second product from the first:
(2x2−x−6) −(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Thus, we have:
f′(x) = x2−4x−2
(x−2)2
Conclusion:
The derivative of the function f(x) = x2+3x−4
x−2using the quotient
rule is:
f′(x) = x2−4x−2
(x−2)2
This result provides the rate of change of the function at any
point x= 2, where the function is defined. Question 4: Quotient Rule
Application for Liberty University Calculus
Problem Statement: Calculate the derivative of the function f(x) =
x2+3x−4
x−2using the quotient rule.
—
Step-by-Step Solution:
Step 1: Identify the Numerator and Denominator
From the function f(x) = x2+3x−4
x−2: - Numerator u(x) = x2+ 3x−4-
Denominator v(x) = x−2
Step 2: Differentiate the Numerator and Denominator
Differentiate u(x):
u′(x) = d
dx (x2+ 3x−4) = 2x+ 3
Differentiate v(x):
v′(x) = d
dx (x−2) = 1
6
Step 3: Apply the Quotient Rule
The quotient rule states:
u
v′
=u′v−uv′
v2
Substituting in the derivatives:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the Expression
Expand the terms in the numerator:
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Now, subtract the second product from the first:
(2x2−x−6) −(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Thus, we have:
f′(x) = x2−4x−2
(x−2)2
Conclusion:
The derivative of the function f(x) = x2+3x−4
x−2using the quotient
rule is:
f′(x) = x2−4x−2
(x−2)2
This result provides the rate of change of the function at any point
x= 2, where the function is defined.
Question 5
Given the functions f(x) = e2xand g(x) = cos(x), find the derivative
of the quotient f(x)
g(x)using the Quotient Rule.
—
Solution:
Step 1: Identify f(x)and g(x)
You are given f(x) = e2xand g(x) = cos(x).
Step 2: Find f′(x)and g′(x)
Differentiate f(x):
f′(x) = d
dx [e2x]=2e2x
7
Differentiate g(x):
g′(x) = d
dx [cos(x)] = −sin(x)
Step 3: Apply the Quotient Rule
The Quotient Rule states:
f
g′
=f′g−fg′
g2
Substitute f(x),g(x),f′(x), and g′(x)into the formula:
e2x
cos(x)′
=(2e2x)(cos(x)) −(e2x)(−sin(x))
(cos(x))2
Step 4: Simplify the expression
Combine terms in the numerator:
=2e2xcos(x) + e2xsin(x)
cos2(x)
Extract common factor e2xin the numerator:
=e2x(2 cos(x) + sin(x))
cos2(x)
Step 5: Final answer
The derivative of the given function e2x
cos(x)is:
e2x
cos(x)′
=e2x(2 cos(x) + sin(x))
cos2(x)
This provides a step-by-step solution to finding the derivative using
the Quotient Rule. Question 5: Quotient Rule Application
Given the functions f(x) = e2xand g(x) = cos(x), find the derivative
of the quotient f(x)
g(x)using the Quotient Rule.
—
Solution:
Step 1: Identify f(x)and g(x)
You are given f(x) = e2xand g(x) = cos(x).
Step 2: Find f′(x)and g′(x)
Differentiate f(x):
f′(x) = d
dx [e2x]=2e2x
Differentiate g(x):
g′(x) = d
dx [cos(x)] = −sin(x)
8
Step 3: Apply the Quotient Rule
The Quotient Rule states:
f
g′
=f′g−fg′
g2
Substitute f(x),g(x),f′(x), and g′(x)into the formula:
e2x
cos(x)′
=(2e2x)(cos(x)) −(e2x)(−sin(x))
(cos(x))2
Step 4: Simplify the expression
Combine terms in the numerator:
=2e2xcos(x) + e2xsin(x)
cos2(x)
Extract common factor e2xin the numerator:
=e2x(2 cos(x) + sin(x))
cos2(x)
Step 5: Final answer
The derivative of the given function e2x
cos(x)is:
e2x
cos(x)′
=e2x(2 cos(x) + sin(x))
cos2(x)
This provides a step-by-step solution to finding the derivative using
the Quotient Rule.
Question 6
Given the function f(x) = x2+3x−4
x−2, use the Quotient Rule to find
f′(x).
Solution
Step 1: Identify the functions u(x)and v(x)To apply the Quotient
Rule, we need to identify the numerator and denominator functions.
Let u(x) = x2+ 3x−4and v(x) = x−2.
Step 2: Differentiate u(x)and v(x)The derivatives of u(x)and v(x)
are calculated as follows:
u′(x) = 2x+ 3
v′(x)=1
Step 3: Apply the Quotient Rule The Quotient Rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
9
Substitute the derivatives and functions into the formula:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the numerator Expand and simplify the expres-
sion in the numerator.
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
(x2+ 3x−4)(1) = x2+ 3x−4
Subtract the second expanded product from the first:
2x2−x−6−(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Step 5: Write the final derivative expression
f′(x) = x2−4x−2
(x−2)2
This is the derivative of the function f(x) = x2+3x−4
x−2using the
Quotient Rule. Question 6
Given the function f(x) = x2+3x−4
x−2, use the Quotient Rule to find
f′(x).
Solution
Step 1: Identify the functions u(x)and v(x)To apply the Quotient
Rule, we need to identify the numerator and denominator functions.
Let u(x) = x2+ 3x−4and v(x) = x−2.
Step 2: Differentiate u(x)and v(x)The derivatives of u(x)and v(x)
are calculated as follows:
u′(x) = 2x+ 3
v′(x)=1
Step 3: Apply the Quotient Rule The Quotient Rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
Substitute the derivatives and functions into the formula:
f′(x) = (2x+ 3)(x−2) −(x2+ 3x−4)(1)
(x−2)2
Step 4: Simplify the numerator Expand and simplify the expres-
sion in the numerator.
(2x+ 3)(x−2) = 2x2−4x+ 3x−6=2x2−x−6
10
(x2+ 3x−4)(1) = x2+ 3x−4
Subtract the second expanded product from the first:
2x2−x−6−(x2+ 3x−4) = 2x2−x−6−x2−3x+ 4 = x2−4x−2
Step 5: Write the final derivative expression
f′(x) = x2−4x−2
(x−2)2
This is the derivative of the function f(x) = x2+3x−4
x−2using the
Quotient Rule.
Question 7
Given the function f(x) = x3+3x2+5
x2, use the quotient rule to find
the derivative f′(x).
Solution:
Step 1: Identify the functions Let u(x) = x3+ 3x2+ 5 and v(x) = x2.
We need to find the derivative using the quotient rule.
Step 2: Apply the Quotient Rule The quotient rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
First, find u′(x)and v′(x):
-u′(x) = d
dx (x3+ 3x2+ 5) = 3x2+ 6x-v′(x) = d
dx (x2)=2x
Step 3: Plug into the formula Plugging these into the quotient
rule gives:
f′(x) = (3x2+ 6x)(x2)−(x3+ 3x2+ 5)(2x)
(x2)2
Simplify the expression inside the numerator:
f′(x) = 3x4+ 6x3−2x4−6x3−10x
x4
f′(x) = x4−10x
x4
Step 4: Simplify the derivative The derivative can be simplified
further:
f′(x) = x−3−10x−3
f′(x) = −9x−3
Step 5: Write in a more common form
f′(x) = −9
x3
11
This is the derivative of the function f(x)using the quotient rule.
Question 7: Quotient Rule Application
Given the function f(x) = x3+3x2+5
x2, use the quotient rule to find
the derivative f′(x).
Solution:
Step 1: Identify the functions Let u(x) = x3+ 3x2+ 5 and v(x) = x2.
We need to find the derivative using the quotient rule.
Step 2: Apply the Quotient Rule The quotient rule states that if
f(x) = u(x)
v(x), then
f′(x) = u′(x)v(x)−u(x)v′(x)
[v(x)]2
First, find u′(x)and v′(x):
-u′(x) = d
dx (x3+ 3x2+ 5) = 3x2+ 6x-v′(x) = d
dx (x2)=2x
Step 3: Plug into the formula Plugging these into the quotient
rule gives:
f′(x) = (3x2+ 6x)(x2)−(x3+ 3x2+ 5)(2x)
(x2)2
Simplify the expression inside the numerator:
f′(x) = 3x4+ 6x3−2x4−6x3−10x
x4
f′(x) = x4−10x
x4
Step 4: Simplify the derivative The derivative can be simplified
further:
f′(x) = x−3−10x−3
f′(x) = −9x−3
Step 5: Write in a more common form
f′(x) = −9
x3
This is the derivative of the function f(x)using the quotient rule.
Question 8
Calculate the derivative of the function f(x) = x3+2x
x2−5.
Solution:
To find the derivative of the given function, we use the Quotient
Rule. The Quotient Rule states that if you have a function defined
as f(x) = g(x)
h(x), then its derivative f′(x)is given by:
f′(x) = g′(x)h(x)−g(x)h′(x)
[h(x)]2
12
Step 1: Identify g(x)and h(x).
In the function f(x) = x3+2x
x2−5: - g(x) = x3+ 2x-h(x) = x2−5
Step 2: Differentiate g(x)and h(x).
- Derivative of g(x) = x3+ 2x:
g′(x)=3x2+ 2
- Derivative of h(x) = x2−5:
h′(x)=2x
Step 3: Apply the Quotient Rule.
Plug in g(x),g′(x),h(x), and h′(x)into the Quotient Rule formula:
f′(x) = (3x2+ 2)(x2−5) −(x3+ 2x)(2x)
(x2−5)2
f′(x) = 3x4−15x2+ 2x2−10 −2x4−4x
(x2−5)2
f′(x) = x4−17x2−4x−10
(x2−5)2
Step 4: Simplify the expression (if possible).
In this case, the expression for f′(x)is already in a relatively sim-
plified form:
f′(x) = x4−17x2−4x−10
(x2−5)2
By following these steps using the Quotient Rule, we have success-
fully calculated the derivative for the function f(x) = x3+2x
x2−5. Question
8: Applying the Quotient Rule
Calculate the derivative of the function f(x) = x3+2x
x2−5.
Solution:
To find the derivative of the given function, we use the Quotient
Rule. The Quotient Rule states that if you have a function defined
as f(x) = g(x)
h(x), then its derivative f′(x)is given by:
f′(x) = g′(x)h(x)−g(x)h′(x)
[h(x)]2
Step 1: Identify g(x)and h(x).
In the function f(x) = x3+2x
x2−5: - g(x) = x3+ 2x-h(x) = x2−5
Step 2: Differentiate g(x)and h(x).
- Derivative of g(x) = x3+ 2x:
g′(x)=3x2+ 2
13
- Derivative of h(x) = x2−5:
h′(x)=2x
Step 3: Apply the Quotient Rule.
Plug in g(x),g′(x),h(x), and h′(x)into the Quotient Rule formula:
f′(x) = (3x2+ 2)(x2−5) −(x3+ 2x)(2x)
(x2−5)2
f′(x) = 3x4−15x2+ 2x2−10 −2x4−4x
(x2−5)2
f′(x) = x4−17x2−4x−10
(x2−5)2
Step 4: Simplify the expression (if possible).
In this case, the expression for f′(x)is already in a relatively sim-
plified form:
f′(x) = x4−17x2−4x−10
(x2−5)2
By following these steps using the Quotient Rule, we have success-
fully calculated the derivative for the function f(x) = x3+2x
x2−5.
Question 9
Problem:
Given the functions f(x) = x3−4x2+ 6x−2and g(x) = x−2, use the
quotient rule to find the derivative of the function h(x) = f(x)
g(x).
—
Solution:
Step 1: Identify the functions involved in the quotient rule
We have: - f(x) = x3−4x2+ 6x−2-g(x) = x−2
Step 2: Write down the formula for the quotient rule
The quotient rule states that if h(x) = f(x)
g(x), then the derivative
h′(x)is given by:
h′(x) = f′(x)g(x)−f(x)g′(x)
[g(x)]2
Step 3: Find the derivatives f′(x)and g′(x)
- Derivative of f(x) = x3−4x2+ 6x−2:
f′(x)=3x2−8x+ 6
- Derivative of g(x) = x−2:
g′(x) = 1
14
Step 4: Substitute f(x),g(x),f′(x), and g′(x)into the quotient rule
formula
h′(x) = (3x2−8x+ 6)(x−2) −(x3−4x2+ 6x−2)(1)
(x−2)2
Step 5: Simplify the numerator
Multiply out the terms in the numerator:
(3x2−8x+ 6)(x−2) = 3x3−6x2−8x2+ 16x+ 6x−12
= 3x3−14x2+ 22x−12
Subtract f(x)multiplied by g′(x):
3x3−14x2+ 22x−12 −(x3−4x2+ 6x−2)
= 3x3−14x2+ 22x−12 −x3+ 4x2−6x+ 2
= 2x3−10x2+ 16x−10
Step 6: Write down the final expression for h′(x)
h′(x) = 2x3−10x2+ 16x−10
(x−2)2
Thus, the derivative of the function h(x) = f(x)
g(x)is h′(x) = 2x3−10x2+16x−10
(x−2)2.
Question 9: Quotient Rule Application
Problem:
Given the functions f(x) = x3−4x2+ 6x−2and g(x) = x−2, use the
quotient rule to find the derivative of the function h(x) = f(x)
g(x).
—
Solution:
Step 1: Identify the functions involved in the quotient rule
We have: - f(x) = x3−4x2+ 6x−2-g(x) = x−2
Step 2: Write down the formula for the quotient rule
The quotient rule states that if h(x) = f(x)
g(x), then the derivative
h′(x)is given by:
h′(x) = f′(x)g(x)−f(x)g′(x)
[g(x)]2
Step 3: Find the derivatives f′(x)and g′(x)
- Derivative of f(x) = x3−4x2+ 6x−2:
f′(x)=3x2−8x+ 6
- Derivative of g(x) = x−2:
g′(x) = 1
15
Step 4: Substitute f(x),g(x),f′(x), and g′(x)into the quotient rule
formula
h′(x) = (3x2−8x+ 6)(x−2) −(x3−4x2+ 6x−2)(1)
(x−2)2
Step 5: Simplify the numerator
Multiply out the terms in the numerator:
(3x2−8x+ 6)(x−2) = 3x3−6x2−8x2+ 16x+ 6x−12
= 3x3−14x2+ 22x−12
Subtract f(x)multiplied by g′(x):
3x3−14x2+ 22x−12 −(x3−4x2+ 6x−2)
= 3x3−14x2+ 22x−12 −x3+ 4x2−6x+ 2
= 2x3−10x2+ 16x−10
Step 6: Write down the final expression for h′(x)
h′(x) = 2x3−10x2+ 16x−10
(x−2)2
Thus, the derivative of the function h(x) = f(x)
g(x)is h′(x) = 2x3−10x2+16x−10
(x−2)2.
Question 10
Problem: Given the functions f(x) = x2+ 5x+ 3 and g(x)=2x+ 1,
find the derivative d
dx f(x)
g(x).
Solution, Step-by-Step:
Step 1: Identify the functions f(x)and g(x). - f(x) = x2+ 5x+ 3 -
g(x)=2x+ 1
Step 2: Compute the derivatives f′(x)and g′(via). - For f(x):
f′(x) = d
dx (x2+ 5x+ 3) = 2x+ 5
- For g(x):
g′(x) = d
dx (2x+ 1) = 2
Step 3: Apply the quotient rule. The quotient rule states that
u
v′=u′v−uv′
v2, where u=f(x)and v=g(x).
- Plug f(x),g(x),f′(x), and g′(x)into the formula:
f(x)
g(x)′
=(2x+ 5)(2x+ 1) −(x2+ 5x+ 3)(2)
(2x+ 1)2
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Step 4: Simplify the numerator. - Expand and simplify:
(2x+ 5)(2x+ 1) = 4x2+ 2x+ 10x+ 5 = 4x2+ 12x+ 5
(x2+ 5x+ 3)(2) = 2x2+ 10x+ 6
4x2+ 12x+ 5 −(2x2+ 10x+ 6) = 4x2+ 12x+ 5 −2x2−10x−6
= 2x2+ 2x−1
Step 5: Write the final derivative expression. - Put the simplified
expression over the squared denominator:
f(x)
g(x)′
=2x2+ 2x−1
(2x+ 1)2
This is the derivative of the function f(x)
g(x)using the quotient rule.
Question 10: Quotient Rule
Problem: Given the functions f(x) = x2+ 5x+ 3 and g(x)=2x+ 1,
find the derivative d
dx f(x)
g(x).
Solution, Step-by-Step:
Step 1: Identify the functions f(x)and g(x). - f(x) = x2+ 5x+ 3 -
g(x)=2x+ 1
Step 2: Compute the derivatives f′(x)and g′(via). - For f(x):
f′(x) = d
dx (x2+ 5x+ 3) = 2x+ 5
- For g(x):
g′(x) = d
dx (2x+ 1) = 2
Step 3: Apply the quotient rule. The quotient rule states that
u
v′=u′v−uv′
v2, where u=f(x)and v=g(x).
- Plug f(x),g(x),f′(x), and g′(x)into the formula:
f(x)
g(x)′
=(2x+ 5)(2x+ 1) −(x2+ 5x+ 3)(2)
(2x+ 1)2
Step 4: Simplify the numerator. - Expand and simplify:
(2x+ 5)(2x+ 1) = 4x2+ 2x+ 10x+ 5 = 4x2+ 12x+ 5
(x2+ 5x+ 3)(2) = 2x2+ 10x+ 6
4x2+ 12x+ 5 −(2x2+ 10x+ 6) = 4x2+ 12x+ 5 −2x2−10x−6
= 2x2+ 2x−1
Step 5: Write the final derivative expression. - Put the simplified
expression over the squared denominator:
f(x)
g(x)′
=2x2+ 2x−1
(2x+ 1)2
This is the derivative of the function f(x)
g(x)using the quotient rule.
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