MATH 100 - FUNDAMENTALS OF
MATHEMATICS - Calculus Question Bank - Set
3
Question 1
Question 1: Find the derivative of the function f(x)=3x2+ 2x−5.
Solution: To find the derivative of the function f(x)=3x2+ 2x−5, we
apply the power rule for derivatives. The power rule states that the derivative
of xnwith respect to xis nxn−1.
Given function: f(x)=3x2+ 2x−5
Step 1: Find the derivative of 3x2
d
dx (3x2)=2·3x2−1= 6x
Step 2: Find the derivative of 2x
d
dx (2x)=2·1x1−1= 2
Step 3: Find the derivative of the constant term −5
d
dx (−5) = 0
Therefore, the derivative of the function f(x)=3x2+ 2x−5 is:
f′(x)=6x+ 2
The derivative of the function is f′(x)=6x+ 2.Certainly! Here is a
question and its solution on Calculus for Liberty University in LateX
format:
Question 1: Find the derivative of the function f(x) = 3x2+ 2x−5.
Solution: To find the derivative of the function f(x) = 3x2+ 2x−5,
we apply the power rule for derivatives. The power rule states that
the derivative of xnwith respect to xis nxn−1.
Given function: f(x)=3x2+ 2x−5
1
Step 1: Find the derivative of 3x2
d
dx (3x2)=2·3x2−1= 6x
Step 2: Find the derivative of 2x
d
dx (2x)=2·1x1−1= 2
Step 3: Find the derivative of the constant term −5
d
dx (−5) = 0
Therefore, the derivative of the function f(x)=3x2+ 2x−5is:
f′(x)=6x+ 2
The derivative of the function is f′(x)=6x+ 2.
Question 2
Solution: To find the derivative of the function f(x) = 3x2−2x+ 5,
we will use the power rule of differentiation.
Given function: f(x)=3x2−2x+ 5
Step 1: Find the derivative of the term 3x2.
d
dx (3x2)=2·3x2−1= 6x
Step 2: Find the derivative of the term −2x.
d
dx (−2x) = −2·1x1−1=−2
Step 3: Find the derivative of the constant term 5. Since the
derivative of a constant is zero,
d
dx 5=0
Therefore, the derivative of the function f(x)=3x2−2x+ 5 is:
f′(x)=6x−2
Question 2: Find the derivative of the following function: f(x) =
3x2−2x+ 5.
Solution: To find the derivative of the function f(x) = 3x2−2x+ 5,
we will use the power rule of differentiation.
Given function: f(x)=3x2−2x+ 5
2
Step 1: Find the derivative of the term 3x2.
d
dx (3x2)=2·3x2−1= 6x
Step 2: Find the derivative of the term −2x.
d
dx (−2x) = −2·1x1−1=−2
Step 3: Find the derivative of the constant term 5. Since the
derivative of a constant is zero,
d
dx 5=0
Therefore, the derivative of the function f(x)=3x2−2x+ 5 is:
f′(x)=6x−2
Question 3
Question 3: Find the derivative of the following function: f(x) =
3x2−5x+ 2
Solution: To find the derivative of the function f(x) = 3x2−5x+ 2,
we will use the power rule for differentiation.
Given function: f(x)=3x2−5x+ 2
Step 1: Differentiate each term of the function separately.
f′(x) = d
dx (3x2)−d
dx (5x) + d
dx (2)
Step 2: Apply the power rule for differentiation.
f′(x)=2·3x2−1−5·1·x1−1+ 0
f′(x)=6x−5
Therefore, the derivative of the function f(x) = 3x2−5x+ 2 is
f′(x) = 6x−5.
3
Sure! Here is a Calculus question along with its step-by-step so-
lution in LaTeX code:
Question 3: Find the derivative of the following function: f(x) =
3x2−5x+ 2
Solution: To find the derivative of the function f(x) = 3x2−5x+ 2,
we will use the power rule for differentiation.
Given function: f(x)=3x2−5x+ 2
Step 1: Differentiate each term of the function separately.
f′(x) = d
dx (3x2)−d
dx (5x) + d
dx (2)
Step 2: Apply the power rule for differentiation.
f′(x)=2·3x2−1−5·1·x1−1+ 0
f′(x)=6x−5
Therefore, the derivative of the function f(x) = 3x2−5x+ 2 is
f′(x) = 6x−5.
4
Question 4
Step-by-step Solution: To find the derivative of the function f(x) =
3x4+ 2x3−5x2+ 7, we will differentiate each term with respect to x
using the power rule.
Given: f(x)=3x4+ 2x3−5x2+ 7
1. Find the derivative of 3x4:d
dx (3x4) = 12x3
2. Find the derivative of 2x3:d
dx (2x3)=6x2
3. Find the derivative of −5x2:d
dx (−5x2) = −10x
4. Find the derivative of the constant term 7: Since the derivative
of a constant is zero, d
dx (7) = 0
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is:
f′(x) = 12x3+ 6x2−10x+ 0
f′(x) = 12x3+ 6x2−10x
Question 4: Find the derivative of the function f(x) = 3x4+2x3−5x2+7.
Step-by-step Solution: To find the derivative of the function f(x) =
3x4+ 2x3−5x2+ 7, we will differentiate each term with respect to x
using the power rule.
Given: f(x)=3x4+ 2x3−5x2+ 7
1. Find the derivative of 3x4:d
dx (3x4) = 12x3
2. Find the derivative of 2x3:d
dx (2x3)=6x2
3. Find the derivative of −5x2:d
dx (−5x2) = −10x
4. Find the derivative of the constant term 7: Since the derivative
of a constant is zero, d
dx (7) = 0
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is:
f′(x) = 12x3+ 6x2−10x+ 0
f′(x) = 12x3+ 6x2−10x
Question 5
Question 5: Find the derivative of the function f(x) = x3−4x2+
2x−7.
Step-by-step solution:
f(x) = x3−4x2+ 2x−7
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (2x)−d
dx (7)
f′(x)=3x2−8x+ 2.
Therefore, the derivative of the function f(x) = x3−4x2+ 2x−7
is f′(x)=3x2−8x+ 2.Certainly! Here’s a question on Calculus along
with the step-by-step solutions in LaTeX code:
5
Question 5: Find the derivative of the function f(x) = x3−4x2+
2x−7.
Step-by-step solution:
f(x) = x3−4x2+ 2x−7
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (2x)−d
dx (7)
f′(x)=3x2−8x+ 2.
Therefore, the derivative of the function f(x) = x3−4x2+ 2x−7is
f′(x) = 3x2−8x+ 2.
Question 6
Question 6:
Find the derivative of the following function:
f(x) = 3x2−2x+ 5
4x−1
Solution:
To find the derivative of the given function f(x) = 3x2−2x+5
4x−1, we can
use the Quotient Rule, which states that if f(x) = g(x)
h(x), then
f′(x) = g′(x)h(x)−g(x)h′(x)
(h(x))2
where g(x)=3x2−2x+ 5 and h(x) = 4x−1.
Now, let’s find the derivative:
g′(x) = 6x−2
h′(x)=4
Plugging these values into the Quotient Rule formula, we get:
f′(x) = (6x−2)(4x−1) −(3x2−2x+ 5) ·4
(4x−1)2
f′(x) = (24x2−6x−8x+ 2) −(12x2−8x+ 20)
(4x−1)2
f′(x) = 12x2−14x−18
(4x−1)2
Therefore, the derivative of the function is:
6
f′(x) = 12x2−14x−18
(4x−1)2
This concludes the solution.
Feel free to reach out if you need any more assistance!Certainly!
Here’s a Calculus question and solution in LateX code:
Question 6:
Find the derivative of the following function:
f(x) = 3x2−2x+ 5
4x−1
Solution:
To find the derivative of the given function f(x) = 3x2−2x+5
4x−1, we can
use the Quotient Rule, which states that if f(x) = g(x)
h(x), then
f′(x) = g′(x)h(x)−g(x)h′(x)
(h(x))2
where g(x)=3x2−2x+ 5 and h(x) = 4x−1.
Now, let’s find the derivative:
g′(x) = 6x−2
h′(x)=4
Plugging these values into the Quotient Rule formula, we get:
f′(x) = (6x−2)(4x−1) −(3x2−2x+ 5) ·4
(4x−1)2
f′(x) = (24x2−6x−8x+ 2) −(12x2−8x+ 20)
(4x−1)2
f′(x) = 12x2−14x−18
(4x−1)2
Therefore, the derivative of the function is:
f′(x) = 12x2−14x−18
(4x−1)2
This concludes the solution.
Feel free to reach out if you need any more assistance!
7
Question 7
Question 7: Find the derivative of the following function: f(x) =
3x2−2x+ 5.
Solution: To find the derivative of the function f(x)=3x2−2x+ 5,
we will use the power rule for differentiation.
Step 1: Identify the terms in the function:
f(x)=3x2−2x+ 5
Step 2: Apply the power rule for differentiation:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
Step 3: Differentiate each term using the power rule:
f′(x)=3·2x2−1−2·1x1−1+ 0
Step 4: Simplify the derivatives:
f′(x)=6x−2
Therefore, the derivative of the function f(x) = 3x2−2x+ 5 is
f′(x) = 6x−2.
f′(x)=6x−2
Certainly! Here is a calculus question along with its step-by-step
solution in LateX code:
Question 7: Find the derivative of the following function: f(x) =
3x2−2x+ 5.
Solution: To find the derivative of the function f(x)=3x2−2x+ 5,
we will use the power rule for differentiation.
Step 1: Identify the terms in the function:
f(x)=3x2−2x+ 5
Step 2: Apply the power rule for differentiation:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
Step 3: Differentiate each term using the power rule:
f′(x)=3·2x2−1−2·1x1−1+ 0
Step 4: Simplify the derivatives:
f′(x)=6x−2
Therefore, the derivative of the function f(x) = 3x2−2x+ 5 is
f′(x) = 6x−2.
f′(x)=6x−2
8
Question 8
Question 8:
Find the derivative of the function f(x) = 3x4−2x3+ 5x−7.
Solution:
To find the derivative of the function f(x) = 3x4−2x3+ 5x−7, we
need to differentiate each term with respect to x.
f′(x) = d
dx [3x4]−d
dx [2x3] + d
dx [5x]−d
dx [7]
Now, taking the derivatives term by term:
f′(x)=4·3x4−1−3·2x3−1+ 5 ·1−0
f′(x) = 12x3−6x2+ 5
Therefore, the derivative of the function f(x)=3x4−2x3+ 5x−7is
f′(x) = 12x3−6x2+ 5.
This is the solution to question 8.Certainly! Here’s a Calculus
question along with its solution in LateX code:
Question 8:
Find the derivative of the function f(x) = 3x4−2x3+ 5x−7.
Solution:
To find the derivative of the function f(x) = 3x4−2x3+ 5x−7, we
need to differentiate each term with respect to x.
f′(x) = d
dx [3x4]−d
dx [2x3] + d
dx [5x]−d
dx [7]
Now, taking the derivatives term by term:
f′(x)=4·3x4−1−3·2x3−1+ 5 ·1−0
f′(x) = 12x3−6x2+ 5
Therefore, the derivative of the function f(x)=3x4−2x3+ 5x−7is
f′(x) = 12x3−6x2+ 5.
This is the solution to question 8.
Question 9
Question 9: Find the derivative of the function f(x)=3x4−2x3+
5x2−7x+ 1.
Solution:
Given function: f(x)=3x4−2x3+ 5x2−7x+ 1
To find the derivative of the function, we need to differentiate each
term separately using the power rule of differentiation.
9
The power rule states that if f(x) = xn, then f′(x) = nxn−1.
Now, let’s differentiate each term of the function:
f′(x) = d
dx (3x4)−d
dx (2x3) + d
dx (5x2)−d
dx (7x) + d
dx (1)
f′(x) = 12x3−6x2+ 10x−7+0
Therefore, the derivative of the function f(x)=3x4−2x3+5x2−7x+1
is f′(x) = 12x3−6x2+ 10x−7.
f′(x) = 12x3−6x2+ 10x−7Certainly! Here’s a question and its
step-by-step solution on Calculus for Liberty University in LateX
code.
Question 9: Find the derivative of the function f(x)=3x4−2x3+
5x2−7x+ 1.
Solution:
Given function: f(x)=3x4−2x3+ 5x2−7x+ 1
To find the derivative of the function, we need to differentiate each
term separately using the power rule of differentiation.
The power rule states that if f(x) = xn, then f′(x) = nxn−1.
Now, let’s differentiate each term of the function:
f′(x) = d
dx (3x4)−d
dx (2x3) + d
dx (5x2)−d
dx (7x) + d
dx (1)
f′(x) = 12x3−6x2+ 10x−7+0
Therefore, the derivative of the function f(x)=3x4−2x3+5x2−7x+1
is f′(x) = 12x3−6x2+ 10x−7.
f′(x) = 12x3−6x2+ 10x−7
Question 10
Question 10: Find the derivative of the function f(x)=4x3−2x2+
5x−1.
Step-by-step Solution: Given function: f(x) = 4x3−2x2+ 5x−1.
To find the derivative of the function, differentiate each term with
respect to xusing the power rule for differentiation:
f′(x) = d
dx (4x3)−d
dx (2x2) + d
dx (5x)−d
dx (1)
= 4 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
= 12x2−4x+ 5.
Therefore, the derivative of the function f(x)=4x3−2x2+ 5x−1is
f′(x) = 12x2−4x+ 5.
This is the required solution to question 10.Sure, here is a question
on Calculus along with its step-by-step solution in LateX code:
Question 10: Find the derivative of the function f(x)=4x3−2x2+
5x−1.
Step-by-step Solution: Given function: f(x) = 4x3−2x2+ 5x−1.
To find the derivative of the function, differentiate each term with
respect to xusing the power rule for differentiation:
10
f′(x) = d
dx (4x3)−d
dx (2x2) + d
dx (5x)−d
dx (1)
= 4 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
= 12x2−4x+ 5.
Therefore, the derivative of the function f(x)=4x3−2x2+ 5x−1is
f′(x) = 12x2−4x+ 5.
This is the required solution to question 10.
Question 11
Question 11: Consider the function f(x) = x3−3x2−9x+ 5. Find
the critical points and determine where the function is increasing or
decreasing.
Solution:
To find the critical points, we first need to find the derivative of
the function f(x).
Given: f(x) = x3−3x2−9x+ 5
The derivative of f(x)is found by differentiating each term sepa-
rately:
f′(x)=3x2−6x−9
Now, to find the critical points, we set the derivative equal to zero
and solve for x:
3x2−6x−9=0
x2−2x−3 = 0
We can factor the quadratic equation:
(x−3)(x+ 1) = 0
So, the critical points are x= 3 and x=−1.
To determine where the function is increasing or decreasing, we
analyze the sign of the derivative f′(x)in the intervals defined by the
critical points.
1. For x < −1: Pick a test point x=−2and substitute into f′(x):
f′(−2) = 3(−2)2−6(−2) −9 = 3(4) + 12 −9 = 15 >0
Therefore, the function is increasing on the interval x < −1.
2. For −1<x<3: Pick a test point x= 0 and substitute into f′(x):
f′(0) = 3(0)2−6(0) −9 = −9<0
Therefore, the function is decreasing on the interval −1<x<3.
11
3. For x > 3: Pick a test point x= 4 and substitute into f′(x):
f′(4) = 3(4)2−6(4) −9 = 3(16) −24 −9 = 27 >0
Therefore, the function is increasing on the interval x > 3.
Therefore, the function is increasing on (−∞,−1) ∪(3,∞)and de-
creasing on (−1,3).
This completes the solution.Sure, here is a Calculus question along
with the step-by-step solution in LateX code:
Question 11: Consider the function f(x) = x3−3x2−9x+ 5. Find
the critical points and determine where the function is increasing or
decreasing.
Solution:
To find the critical points, we first need to find the derivative of
the function f(x).
Given: f(x) = x3−3x2−9x+ 5
The derivative of f(x)is found by differentiating each term sepa-
rately:
f′(x) = 3x2−6x−9
Now, to find the critical points, we set the derivative equal to zero
and solve for x:
3x2−6x−9=0
x2−2x−3 = 0
We can factor the quadratic equation:
(x−3)(x+ 1) = 0
So, the critical points are x= 3 and x=−1.
To determine where the function is increasing or decreasing, we
analyze the sign of the derivative f′(x)in the intervals defined by the
critical points.
1. For x < −1: Pick a test point x=−2and substitute into f′(x):
f′(−2) = 3(−2)2−6(−2) −9 = 3(4) + 12 −9 = 15 >0
Therefore, the function is increasing on the interval x < −1.
2. For −1<x<3: Pick a test point x= 0 and substitute into f′(x):
f′(0) = 3(0)2−6(0) −9 = −9<0
Therefore, the function is decreasing on the interval −1<x<3.
3. For x > 3: Pick a test point x= 4 and substitute into f′(x):
f′(4) = 3(4)2−6(4) −9 = 3(16) −24 −9 = 27 >0
Therefore, the function is increasing on the interval x > 3.
Therefore, the function is increasing on (−∞,−1) ∪(3,∞)and de-
creasing on (−1,3).
This completes the solution.
12
Question 12
Find the antiderivative of the function f(x)=3x2−2x+ 5.
Step-by-step solution:
To find the antiderivative of the given function, we need to inte-
grate each term separately.
Given function: f(x)=3x2−2x+ 5
Integrating with respect to x:
Z3x2dx −Z2xdx +Z5dx
Integrating each term:
=3
3x3−2
2x2+ 5x+C
where Cis the constant of integration.
Simplifying:
=x3−x2+ 5x+C
Therefore, the antiderivative of the function f(x) = 3x2−2x+ 5 is
x3−x2+ 5x+C, where Cis the constant of integration.Question 12:
Find the antiderivative of the function f(x)=3x2−2x+ 5.
Step-by-step solution:
To find the antiderivative of the given function, we need to inte-
grate each term separately.
Given function: f(x)=3x2−2x+ 5
Integrating with respect to x:
Z3x2dx −Z2xdx +Z5dx
Integrating each term:
=3
3x3−2
2x2+ 5x+C
where Cis the constant of integration.
Simplifying:
=x3−x2+ 5x+C
Therefore, the antiderivative of the function f(x) = 3x2−2x+ 5 is
x3−x2+ 5x+C, where Cis the constant of integration.
13
Question 13
Question 13: Find the derivative of the function f(x) = x3−4x2+
5x−2.
Solution: To find the derivative of f(x) = x3−4x2+ 5x−2, we will
use the power rule for differentiation.
Given function, f(x) = x3−4x2+ 5x−2
Derivative of f(x)is given by:
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (5x)−d
dx (2)
Computing the derivatives:
f′(x) = 3x2−8x+ 5 −0
Therefore, the derivative of the function f(x) = x3−4x2+ 5x−2is
the function:
f′(x) = 3x2−8x+ 5
Absolutely! Here is a Calculus question along with step-by-step so-
lutions in LaTeX code:
Question 13: Find the derivative of the function f(x) = x3−4x2+
5x−2.
Solution: To find the derivative of f(x) = x3−4x2+ 5x−2, we will
use the power rule for differentiation.
Given function, f(x) = x3−4x2+ 5x−2
Derivative of f(x)is given by:
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (5x)−d
dx (2)
Computing the derivatives:
f′(x) = 3x2−8x+ 5 −0
Therefore, the derivative of the function f(x) = x3−4x2+ 5x−2is
the function:
f′(x) = 3x2−8x+ 5
Question 14
Question 14: Find the derivative of the following function:
f(x)=3x3−2x2+ 5x−7
Solution: To find the derivative of f(x) = 3x3−2x2+ 5x−7, we will
differentiate each term separately using the power rule.
14
f′(x) = d
dx 3x3−d
dx 2x2+d
dx (5x)−d
dx (7)
Applying the power rule:
f′(x) = 3 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying the derivatives:
f′(x) = 9x2−4x+ 5
Therefore, the derivative of the function f(x)=3x3−2x2+ 5x−7is
f′(x) = 9x2−4x+ 5.
This is question 14 for set 3.Certainly! Here is a question on
Calculus for Liberty University along with a step-by-step solution in
LaTeX code:
Question 14: Find the derivative of the following function:
f(x)=3x3−2x2+ 5x−7
Solution: To find the derivative of f(x) = 3x3−2x2+ 5x−7, we will
differentiate each term separately using the power rule.
f′(x) = d
dx 3x3−d
dx 2x2+d
dx (5x)−d
dx (7)
Applying the power rule:
f′(x) = 3 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying the derivatives:
f′(x) = 9x2−4x+ 5
Therefore, the derivative of the function f(x)=3x3−2x2+ 5x−7is
f′(x) = 9x2−4x+ 5.
This is question 14 for set 3.
Question 15
Question 15: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Solution: To find the derivative of the function f(x) = 3x2+ 5x+ 2,
we differentiate each term with respect to x. Using the power rule
for differentiation, we have:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
f′(x)=3·2x2−1+ 5 ·1x1−1+ 0
15
f′(x)=6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.Certainly! Here’s a Calculus question along with its
solution in LateX:
Question 15: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Solution: To find the derivative of the function f(x) = 3x2+ 5x+ 2,
we differentiate each term with respect to x. Using the power rule
for differentiation, we have:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
f′(x)=3·2x2−1+ 5 ·1x1−1+ 0
f′(x)=6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.
Question 16
Question 16: Find the derivative of the function f(x) = e3x2+2x.
Solution: To find the derivative of the given function f(x) = e3x2+2x,
we will use the chain rule.
Step 1: Identify the function and its components. Let u= 3x2+ 2x.
Then, our function can be written as f(x) = eu.
Step 2: Apply the chain rule. The chain rule states that if f(x) =
eu, then d
dx (eu) = eu·du
dx .
Step 3: Find du
dx .
du
dx =d
dx (3x2+ 2x) = 6x+ 2
Step 4: Substitute back into the chain rule formula.
d
dx (e3x2+2x) = e3x2+2x·(6x+ 2)
Therefore, the derivative of the function is f′(x) = e3x2+2x·(6x+
2).Sure, here is a Calculus question along with a step-by-step solution
in LateX format:
Question 16: Find the derivative of the function f(x) = e3x2+2x.
Solution: To find the derivative of the given function f(x) = e3x2+2x,
we will use the chain rule.
16
Step 1: Identify the function and its components. Let u= 3x2+ 2x.
Then, our function can be written as f(x) = eu.
Step 2: Apply the chain rule. The chain rule states that if f(x) =
eu, then d
dx (eu) = eu·du
dx .
Step 3: Find du
dx .
du
dx =d
dx (3x2+ 2x) = 6x+ 2
Step 4: Substitute back into the chain rule formula.
d
dx (e3x2+2x) = e3x2+2x·(6x+ 2)
Therefore, the derivative of the function is f′(x) = e3x2+2x·(6x+ 2).
Question 17
Question 17: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Step-by-step Solution: To find the derivative of the function f(x) =
3x2+ 5x+ 2, we use the power rule.
1. Differentiate each term of the function separately:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
2. Apply the power rule to each term:
f′(x) = 3 ·2x2−1+ 5 ·1x1−1+ 0
3. Simplify the derivatives:
f′(x) = 6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.Sure! Here is a Calculus question along with its step-
by-step solution in LateX code:
Question 17: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Step-by-step Solution: To find the derivative of the function f(x) =
3x2+ 5x+ 2, we use the power rule.
1. Differentiate each term of the function separately:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
2. Apply the power rule to each term:
f′(x) = 3 ·2x2−1+ 5 ·1x1−1+ 0
3. Simplify the derivatives:
f′(x) = 6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.
Question 18
Question 18: Find the derivative of the function f(x)=3x4+ 2x3−
5x2+ 7.
17
Solution: To find the derivative of the function f(x) = 3x4+ 2x3−
5x2+ 7, we will apply the power rule for derivatives to each term.
Given function: f(x)=3x4+ 2x3−5x2+ 7
Step 1: Find the derivative of 3x4
f′(x)=4·3x4−1= 12x3
Step 2: Find the derivative of 2x3
f′(x)=3·2x3−1= 6x2
Step 3: Find the derivative of −5x2
f′(x)=2·(−5)x2−1=−10x
Step 4: Find the derivative of the constant term 7
f′(x)=0
(The derivative of a constant term is 0)
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is
f′(x) = 12x3+ 6x2−10x
Feel free to reach out if you need further assistance or more ques-
tions!Certainly! Here is a Calculus question along with its step-by-
step solution in LateX code:
Question 18: Find the derivative of the function f(x)=3x4+ 2x3−
5x2+ 7.
Solution: To find the derivative of the function f(x) = 3x4+ 2x3−
5x2+ 7, we will apply the power rule for derivatives to each term.
Given function: f(x)=3x4+ 2x3−5x2+ 7
Step 1: Find the derivative of 3x4
f′(x)=4·3x4−1= 12x3
Step 2: Find the derivative of 2x3
f′(x)=3·2x3−1= 6x2
Step 3: Find the derivative of −5x2
f′(x) = 2 ·(−5)x2−1=−10x
Step 4: Find the derivative of the constant term 7
f′(x) = 0
(The derivative of a constant term is 0)
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is
f′(x) = 12x3+ 6x2−10x
Feel free to reach out if you need further assistance or more ques-
tions!
18
Question 19
Question 19: Let f(x) = excos x. Find f′′ (x).
Solution: To find f′′ (x), we first need to find f′(x)and then take
its derivative.
Given: f(x) = excos x
Step 1: Find f′(x)
f′(x) = (excos x)′=ex(−sin x) + cos xex=ex(cos x−sin x)
Step 2: Find f′′ (x)
f′′ (x) = (ex(cos x−sin x))′=ex(−sin x−cos x)+ex(cos x−sin x) = ex(−sin x−cos x+cos x−sin x) = ex(−2 sin x) = −2exsin x
Therefore, the second derivative of f(x) = excos xis f′′ (x) = −2exsin x.Certainly!
Here is a Calculus question along with its step-by-step solution in La-
teX code:
Question 19: Let f(x) = excos x. Find f′′ (x).
Solution: To find f′′ (x), we first need to find f′(x)and then take
its derivative.
Given: f(x) = excos x
Step 1: Find f′(x)
f′(x) = (excos x)′=ex(−sin x) + cos xex=ex(cos x−sin x)
Step 2: Find f′′ (x)
f′′ (x)=(ex(cos x−sin x))′=ex(−sin x−cos x)+ex(cos x−sin x) = ex(−sin x−cos x+cos x−sin x) = ex(−2 sin x) = −2exsin x
Therefore, the second derivative of f(x) = excos xis f′′ (x) = −2exsin x.
Question 20
Step-by-step solution: To find the derivative of f(x)=2x3−5x2+
4x−7, we will differentiate each term with respect to xusing the
power rule.
1. Differentiate 2x3:
d
dx (2x3) = 2 ×3x3−1= 6x2
2. Differentiate −5x2:
d
dx (−5x2) = −5×2x2−1=−10x
3. Differentiate 4x:d
dx (4x)=4
19
4. Differentiate the constant term −7: The derivative of a constant
is always zero: d
dx (−7) = 0
Therefore, the derivative of f(x)=2x3−5x2+ 4x−7is:
f′(x)=6x2−10x+ 4
Question 20: Compute the derivative of the function f(x)=2x3−5x2+
4x−7.
Step-by-step solution: To find the derivative of f(x)=2x3−5x2+
4x−7, we will differentiate each term with respect to xusing the
power rule.
1. Differentiate 2x3:
d
dx (2x3) = 2 ×3x3−1= 6x2
2. Differentiate −5x2:
d
dx (−5x2) = −5×2x2−1=−10x
3. Differentiate 4x:d
dx (4x)=4
4. Differentiate the constant term −7: The derivative of a constant
is always zero: d
dx (−7) = 0
Therefore, the derivative of f(x)=2x3−5x2+ 4x−7is:
f′(x)=6x2−10x+ 4
Question 21
Find the derivative of the following function using the Power Rule:
f(x) = 3x5−2x3+ 4x−6
Solution:
To find the derivative of the function f(x)=3x5−2x3+ 4x−6using
the Power Rule, we differentiate each term separately.
1. Differentiate 3x5:
d
dx (3x5) = 5 ·3x5−1= 15x4
2. Differentiate −2x3:
d
dx (−2x3)=3·(−2)x3−1=−6x2
3. Differentiate 4x:d
dx (4x)=4
20
4. Differentiate −6(a constant):
d
dx (−6) = 0
Therefore, the derivative of the function f(x)=3x5−2x3+ 4x−6is:
f′(x) = 15x4−6x2+ 4
Question 21:
Find the derivative of the following function using the Power Rule:
f(x)=3x5−2x3+ 4x−6
Solution:
To find the derivative of the function f(x)=3x5−2x3+ 4x−6using
the Power Rule, we differentiate each term separately.
1. Differentiate 3x5:
d
dx (3x5) = 5 ·3x5−1= 15x4
2. Differentiate −2x3:
d
dx (−2x3)=3·(−2)x3−1=−6x2
3. Differentiate 4x:d
dx (4x)=4
4. Differentiate −6(a constant):
d
dx (−6) = 0
Therefore, the derivative of the function f(x)=3x5−2x3+ 4x−6is:
f′(x) = 15x4−6x2+ 4
Question 22
Question 22: Find the derivative of the function f(x)=3x4−2x2+5.
Solution: To find the derivative of the function f(x)=3x4−2x2+ 5,
we will differentiate each term with respect to xusing the power rule
of differentiation.
The power rule states that if the function is in the form f(x) = axn,
then the derivative is f′(x) = anxn−1.
Given f(x)=3x4−2x2+ 5, we differentiate each term as follows:
f′(x) = d
dx (3x4)−d
dx (2x2) + d
dx (5).
21
Applying the power rule: f′(x)=3·4x4−1−2·2x2−1+ 0.
Simplify the derivatives: f′(x) = 12x3−4x.
Therefore, the derivative of the function f(x) = 3x4−2x2+ 5 is
f′(x) = 12x3−4x.
I hope this helps! Let me know if you need any more assis-
tance.Certainly! Here’s a calculus question along with its solution
in LateX code:
Question 22: Find the derivative of the function f(x)=3x4−2x2+5.
Solution: To find the derivative of the function f(x)=3x4−2x2+ 5,
we will differentiate each term with respect to xusing the power rule
of differentiation.
The power rule states that if the function is in the form f(x) = axn,
then the derivative is f′(x) = anxn−1.
Given f(x)=3x4−2x2+ 5, we differentiate each term as follows:
f′(x) = d
dx (3x4)−d
dx (2x2) + d
dx (5).
Applying the power rule: f′(x)=3·4x4−1−2·2x2−1+ 0.
Simplify the derivatives: f′(x) = 12x3−4x.
Therefore, the derivative of the function f(x) = 3x4−2x2+ 5 is
f′(x) = 12x3−4x.
I hope this helps! Let me know if you need any more assistance.
Question 23
Question 23:
Find the derivative of the function f(x) = 3x2−2x+ 5.
Solution:
To find the derivative of the function f(x)=3x2−2x+5, we’ll apply
the power rule for differentiation.
Given function: f(x)=3x2−2x+ 5
Step 1: Find the derivative of each term separately.
d
dx (3x2)=2·3x2−1= 6x
d
dx (−2x) = −2
Step 2: Combine the derivatives of individual terms.
The derivative of the function f(x)=3x2−2x+ 5 is:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
f′(x)=6x−2+0
f′(x)=6x−2
22
Therefore, the derivative of the function f(x)=3x2−2x+5 is f′(x) =
6x−2.Sure, here is a question on Calculus for Liberty University along
with its step-by-step solution presented in LateX code:
Question 23:
Find the derivative of the function f(x) = 3x2−2x+ 5.
Solution:
To find the derivative of the function f(x)=3x2−2x+5, we’ll apply
the power rule for differentiation.
Given function: f(x)=3x2−2x+ 5
Step 1: Find the derivative of each term separately.
d
dx (3x2)=2·3x2−1= 6x
d
dx (−2x) = −2
Step 2: Combine the derivatives of individual terms.
The derivative of the function f(x)=3x2−2x+ 5 is:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
f′(x)=6x−2+0
f′(x)=6x−2
Therefore, the derivative of the function f(x) = 3x2−2x+ 5 is
f′(x) = 6x−2.
Question 24
Question 24: Find the derivative of the function f(x)=3x4−8x2+
5x−2.
Solution: To find the derivative of f(x), we can apply the power
rule for differentiation. The power rule states that if f(x) = axn, then
f′(x) = naxn−1.
Given function: f(x)=3x4−8x2+ 5x−2
Applying the power rule:
f′(x) = 4(3)x4−1−2(8)x2−1+ 5x0
Simplifying the derivative:
f′(x) = 12x3−16x+ 5
Therefore, the derivative of the function f(x)=3x4−8x2+ 5x−2is
f′(x) = 12x3−16x+ 5.Sure, here is a calculus question along with its
step-by-step solution in LateX code:
23
Question 24: Find the derivative of the function f(x)=3x4−8x2+
5x−2.
Solution: To find the derivative of f(x), we can apply the power
rule for differentiation. The power rule states that if f(x) = axn, then
f′(x) = naxn−1.
Given function: f(x)=3x4−8x2+ 5x−2
Applying the power rule:
f′(x) = 4(3)x4−1−2(8)x2−1+ 5x0
Simplifying the derivative:
f′(x) = 12x3−16x+ 5
Therefore, the derivative of the function f(x)=3x4−8x2+ 5x−2is
f′(x) = 12x3−16x+ 5.
Question 25
Step 1: Determine the critical points by finding the derivative of
the function:
f′(x)=3x2−6x
Step 2: Set f′(x)=0to find the critical points:
3x2−6x= 0
3x(x−2) = 0
So, the critical points are x= 0 and x= 2.
Step 3: Find the function values at the critical points and end-
points of the interval: - f(0) = 03−3(0)2+ 2 = 2 -f(3) = 33−3(3)2+2 = 2
-f(2) = 23−3(2)2+ 2 = −2
Step 4: Compare the function values at critical points and end-
points to determine the absolute maximum and minimum: - The
maximum value is f(0) = 2 at x= 0 - The minimum value is f(2) = −2
at x= 2
Therefore, the absolute maximum value of the function on the
interval [0,3] is 2 and the absolute minimum value is -2.Question 25:
Find the absolute maximum and minimum values of the function
f(x) = x3−3x2+ 2 on the interval [0,3].
Step 1: Determine the critical points by finding the derivative of
the function:
f′(x)=3x2−6x
Step 2: Set f′(x)=0to find the critical points:
3x2−6x= 0
24
3x(x−2) = 0
So, the critical points are x= 0 and x= 2.
Step 3: Find the function values at the critical points and end-
points of the interval: - f(0) = 03−3(0)2+ 2 = 2 -f(3) = 33−3(3)2+2 = 2
-f(2) = 23−3(2)2+ 2 = −2
Step 4: Compare the function values at critical points and end-
points to determine the absolute maximum and minimum: - The
maximum value is f(0) = 2 at x= 0 - The minimum value is f(2) = −2
at x= 2
Therefore, the absolute maximum value of the function on the
interval [0,3] is 2 and the absolute minimum value is -2.
Question 26
Question 26: Find the derivative of the function f(x)=3x4−2x2+
5x−7.
Step-by-step Solution: To find the derivative of the function f(x) =
3x4−2x2+5x−7, we will use the power rule of differentiation. Applying
the power rule to each term of the function, we get:
f′(x) = 4 ·3x4−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying, we have:
f′(x) = 12x3−4x+ 5
Therefore, the derivative of the function f(x)=3x4−2x2+ 5x−7is
f′(x) = 12x3−4x+ 5.Sure, here is a question on calculus along with its
step-by-step solutions in LateX code:
Question 26: Find the derivative of the function f(x)=3x4−2x2+
5x−7.
Step-by-step Solution: To find the derivative of the function f(x) =
3x4−2x2+5x−7, we will use the power rule of differentiation. Applying
the power rule to each term of the function, we get:
f′(x) = 4 ·3x4−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying, we have:
f′(x) = 12x3−4x+ 5
Therefore, the derivative of the function f(x)=3x4−2x2+ 5x−7is
f′(x) = 12x3−4x+ 5.
25
Question 27
Step 1: Apply the Power Rule to each term of the function.
f′(x) = d
dx (3x4)−d
dx (8x2) + d
dx (5)
= 3 ·4x4−1−8·2x2−1+ 0
= 12x3−16x
Step 2: Simplify the derivative.
f′(x) = 12x3−16x
Therefore, the derivative of the function f(x) = 3x4−8x2+ 5 is
f′(x) = 12x3−16x.Question 27: Find the derivative of the following
function using the Power Rule.
f(x)=3x4−8x2+ 5
Step 1: Apply the Power Rule to each term of the function.
f′(x) = d
dx (3x4)−d
dx (8x2) + d
dx (5)
= 3 ·4x4−1−8·2x2−1+ 0
= 12x3−16x
Step 2: Simplify the derivative.
f′(x) = 12x3−16x
Therefore, the derivative of the function f(x) = 3x4−8x2+ 5 is
f′(x) = 12x3−16x.
Question 28
Question 28: Find the derivative of the function f(x)=3x2+ 4x−5.
Step-by-step Solution: To find the derivative of the function, we
apply the power rule for derivatives. The power rule states that the
derivative of xnis nxn−1.
Given function: f(x)=3x2+ 4x−5
1. Find the derivative of 3x2:
d
dx (3x2)=2·3x2−1= 6x
2. Find the derivative of 4x:
d
dx (4x)=4·x1−1= 4
26
3. Find the derivative of the constant term −5: The derivative of
a constant is 0.
Therefore, the derivative of the function f(x)=3x2+ 4x−5is:
f′(x)=6x+ 4
Certainly! Here is a Calculus question along with the step-by-step
solution presented in LaTeX code:
Question 28: Find the derivative of the function f(x)=3x2+ 4x−5.
Step-by-step Solution: To find the derivative of the function, we
apply the power rule for derivatives. The power rule states that the
derivative of xnis nxn−1.
Given function: f(x)=3x2+ 4x−5
1. Find the derivative of 3x2:
d
dx (3x2)=2·3x2−1= 6x
2. Find the derivative of 4x:
d
dx (4x)=4·x1−1= 4
3. Find the derivative of the constant term −5: The derivative of
a constant is 0.
Therefore, the derivative of the function f(x)=3x2+ 4x−5is:
f′(x)=6x+ 4
Question 29
Question 29: Find the derivative of the function f(x)=5x3−2x2+
4x+ 7.
Step-by-step solution: To find the derivative of the function f(x) =
5x3−2x2+ 4x+ 7, we will differentiate each term using the power rule.
f′(x) = d
dx (5x3)−d
dx (2x2) + d
dx (4x) + d
dx (7)
= 5 ·3x3−1−2·2x2−1+ 4 ·1x1−1+ 0
= 15x2−4x+ 4.
Therefore, the derivative of the function f(x)=5x3−2x2+ 4x+ 7
is f′(x) = 15x2−4x+ 4.Certainly! Here is a question on Calculus
for Liberty University along with the step-by-step solution in LateX
code:
Question 29: Find the derivative of the function f(x)=5x3−2x2+
4x+ 7.
27
Step-by-step solution: To find the derivative of the function f(x) =
5x3−2x2+ 4x+ 7, we will differentiate each term using the power rule.
f′(x) = d
dx (5x3)−d
dx (2x2) + d
dx (4x) + d
dx (7)
= 5 ·3x3−1−2·2x2−1+ 4 ·1x1−1+ 0
= 15x2−4x+ 4.
Therefore, the derivative of the function f(x)=5x3−2x2+ 4x+ 7 is
f′(x) = 15x2−4x+ 4.
Question 30
Question: Calculate the derivative of the function f(x) = x3−2x2+
5x−3.
Step-by-step Solution: To find the derivative of the function f(x) =
x3−2x2+ 5x−3, we will differentiate each term separately using the
power rule for derivatives.
1. Find the derivative of the term x3:
d
dx (x3) = 3x3−1= 3x2
2. Find the derivative of the term −2x2:
d
dx (−2x2) = −2·2x2−1=−4x
3. Find the derivative of the term 5x:
d
dx (5x) = 5 ·1=5
4. Find the derivative of the constant term −3:
d
dx (−3) = 0
Therefore, the derivative of the function f(x) = x3−2x2+ 5x−3is:
f′(x) = 3x2−4x+ 5
This completes the solution.
I hope this helps! Let me know if you need any further assis-
tance.Certainly! Here is a Calculus question along with its step-by-
step solution in LateX code:
Question: Calculate the derivative of the function f(x) = x3−2x2+
5x−3.
28
Step 1: Find the derivative of 3x2
d
dx (3x2)=2·3x2−1= 6x
Step 2: Find the derivative of 2x
d
dx (2x)=2·1x1−1= 2
Step 3: Find the derivative of the constant term −5
d
dx (−5) = 0
Therefore, the derivative of the function f(x)=3x2+ 2x−5is:
f′(x)=6x+ 2
The derivative of the function is f′(x)=6x+ 2.
Question 2
Solution: To find the derivative of the function f(x) = 3x2−2x+ 5,
we will use the power rule of differentiation.
Given function: f(x)=3x2−2x+ 5
Step 1: Find the derivative of the term 3x2.
d
dx (3x2)=2·3x2−1= 6x
Step 2: Find the derivative of the term −2x.
d
dx (−2x) = −2·1x1−1=−2
Step 3: Find the derivative of the constant term 5. Since the
derivative of a constant is zero,
d
dx 5=0
Therefore, the derivative of the function f(x)=3x2−2x+ 5 is:
f′(x)=6x−2
Question 2: Find the derivative of the following function: f(x) =
3x2−2x+ 5.
Solution: To find the derivative of the function f(x) = 3x2−2x+ 5,
we will use the power rule of differentiation.
Given function: f(x)=3x2−2x+ 5
2
Step 1: Find the derivative of the term 3x2.
d
dx (3x2)=2·3x2−1= 6x
Step 2: Find the derivative of the term −2x.
d
dx (−2x) = −2·1x1−1=−2
Step 3: Find the derivative of the constant term 5. Since the
derivative of a constant is zero,
d
dx 5=0
Therefore, the derivative of the function f(x)=3x2−2x+ 5 is:
f′(x)=6x−2
Question 3
Question 3: Find the derivative of the following function: f(x) =
3x2−5x+ 2
Solution: To find the derivative of the function f(x) = 3x2−5x+ 2,
we will use the power rule for differentiation.
Given function: f(x)=3x2−5x+ 2
Step 1: Differentiate each term of the function separately.
f′(x) = d
dx (3x2)−d
dx (5x) + d
dx (2)
Step 2: Apply the power rule for differentiation.
f′(x)=2·3x2−1−5·1·x1−1+ 0
f′(x)=6x−5
Therefore, the derivative of the function f(x) = 3x2−5x+ 2 is
f′(x) = 6x−5.
3
Sure! Here is a Calculus question along with its step-by-step so-
lution in LaTeX code:
Question 3: Find the derivative of the following function: f(x) =
3x2−5x+ 2
Solution: To find the derivative of the function f(x) = 3x2−5x+ 2,
we will use the power rule for differentiation.
Given function: f(x)=3x2−5x+ 2
Step 1: Differentiate each term of the function separately.
f′(x) = d
dx (3x2)−d
dx (5x) + d
dx (2)
Step 2: Apply the power rule for differentiation.
f′(x)=2·3x2−1−5·1·x1−1+ 0
f′(x)=6x−5
Therefore, the derivative of the function f(x) = 3x2−5x+ 2 is
f′(x) = 6x−5.
4
Question 4
Step-by-step Solution: To find the derivative of the function f(x) =
3x4+ 2x3−5x2+ 7, we will differentiate each term with respect to x
using the power rule.
Given: f(x)=3x4+ 2x3−5x2+ 7
1. Find the derivative of 3x4:d
dx (3x4) = 12x3
2. Find the derivative of 2x3:d
dx (2x3)=6x2
3. Find the derivative of −5x2:d
dx (−5x2) = −10x
4. Find the derivative of the constant term 7: Since the derivative
of a constant is zero, d
dx (7) = 0
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is:
f′(x) = 12x3+ 6x2−10x+ 0
f′(x) = 12x3+ 6x2−10x
Question 4: Find the derivative of the function f(x) = 3x4+2x3−5x2+7.
Step-by-step Solution: To find the derivative of the function f(x) =
3x4+ 2x3−5x2+ 7, we will differentiate each term with respect to x
using the power rule.
Given: f(x)=3x4+ 2x3−5x2+ 7
1. Find the derivative of 3x4:d
dx (3x4) = 12x3
2. Find the derivative of 2x3:d
dx (2x3)=6x2
3. Find the derivative of −5x2:d
dx (−5x2) = −10x
4. Find the derivative of the constant term 7: Since the derivative
of a constant is zero, d
dx (7) = 0
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is:
f′(x) = 12x3+ 6x2−10x+ 0
f′(x) = 12x3+ 6x2−10x
Question 5
Question 5: Find the derivative of the function f(x) = x3−4x2+
2x−7.
Step-by-step solution:
f(x) = x3−4x2+ 2x−7
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (2x)−d
dx (7)
f′(x)=3x2−8x+ 2.
Therefore, the derivative of the function f(x) = x3−4x2+ 2x−7
is f′(x)=3x2−8x+ 2.Certainly! Here’s a question on Calculus along
with the step-by-step solutions in LaTeX code:
5
Question 5: Find the derivative of the function f(x) = x3−4x2+
2x−7.
Step-by-step solution:
f(x) = x3−4x2+ 2x−7
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (2x)−d
dx (7)
f′(x)=3x2−8x+ 2.
Therefore, the derivative of the function f(x) = x3−4x2+ 2x−7is
f′(x) = 3x2−8x+ 2.
Question 6
Question 6:
Find the derivative of the following function:
f(x) = 3x2−2x+ 5
4x−1
Solution:
To find the derivative of the given function f(x) = 3x2−2x+5
4x−1, we can
use the Quotient Rule, which states that if f(x) = g(x)
h(x), then
f′(x) = g′(x)h(x)−g(x)h′(x)
(h(x))2
where g(x)=3x2−2x+ 5 and h(x) = 4x−1.
Now, let’s find the derivative:
g′(x) = 6x−2
h′(x)=4
Plugging these values into the Quotient Rule formula, we get:
f′(x) = (6x−2)(4x−1) −(3x2−2x+ 5) ·4
(4x−1)2
f′(x) = (24x2−6x−8x+ 2) −(12x2−8x+ 20)
(4x−1)2
f′(x) = 12x2−14x−18
(4x−1)2
Therefore, the derivative of the function is:
6
f′(x) = 12x2−14x−18
(4x−1)2
This concludes the solution.
Feel free to reach out if you need any more assistance!Certainly!
Here’s a Calculus question and solution in LateX code:
Question 6:
Find the derivative of the following function:
f(x) = 3x2−2x+ 5
4x−1
Solution:
To find the derivative of the given function f(x) = 3x2−2x+5
4x−1, we can
use the Quotient Rule, which states that if f(x) = g(x)
h(x), then
f′(x) = g′(x)h(x)−g(x)h′(x)
(h(x))2
where g(x)=3x2−2x+ 5 and h(x) = 4x−1.
Now, let’s find the derivative:
g′(x) = 6x−2
h′(x)=4
Plugging these values into the Quotient Rule formula, we get:
f′(x) = (6x−2)(4x−1) −(3x2−2x+ 5) ·4
(4x−1)2
f′(x) = (24x2−6x−8x+ 2) −(12x2−8x+ 20)
(4x−1)2
f′(x) = 12x2−14x−18
(4x−1)2
Therefore, the derivative of the function is:
f′(x) = 12x2−14x−18
(4x−1)2
This concludes the solution.
Feel free to reach out if you need any more assistance!
7
Question 7
Question 7: Find the derivative of the following function: f(x) =
3x2−2x+ 5.
Solution: To find the derivative of the function f(x)=3x2−2x+ 5,
we will use the power rule for differentiation.
Step 1: Identify the terms in the function:
f(x)=3x2−2x+ 5
Step 2: Apply the power rule for differentiation:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
Step 3: Differentiate each term using the power rule:
f′(x)=3·2x2−1−2·1x1−1+ 0
Step 4: Simplify the derivatives:
f′(x)=6x−2
Therefore, the derivative of the function f(x) = 3x2−2x+ 5 is
f′(x) = 6x−2.
f′(x)=6x−2
Certainly! Here is a calculus question along with its step-by-step
solution in LateX code:
Question 7: Find the derivative of the following function: f(x) =
3x2−2x+ 5.
Solution: To find the derivative of the function f(x)=3x2−2x+ 5,
we will use the power rule for differentiation.
Step 1: Identify the terms in the function:
f(x)=3x2−2x+ 5
Step 2: Apply the power rule for differentiation:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
Step 3: Differentiate each term using the power rule:
f′(x)=3·2x2−1−2·1x1−1+ 0
Step 4: Simplify the derivatives:
f′(x)=6x−2
Therefore, the derivative of the function f(x) = 3x2−2x+ 5 is
f′(x) = 6x−2.
f′(x)=6x−2
8
Question 8
Question 8:
Find the derivative of the function f(x) = 3x4−2x3+ 5x−7.
Solution:
To find the derivative of the function f(x) = 3x4−2x3+ 5x−7, we
need to differentiate each term with respect to x.
f′(x) = d
dx [3x4]−d
dx [2x3] + d
dx [5x]−d
dx [7]
Now, taking the derivatives term by term:
f′(x)=4·3x4−1−3·2x3−1+ 5 ·1−0
f′(x) = 12x3−6x2+ 5
Therefore, the derivative of the function f(x)=3x4−2x3+ 5x−7is
f′(x) = 12x3−6x2+ 5.
This is the solution to question 8.Certainly! Here’s a Calculus
question along with its solution in LateX code:
Question 8:
Find the derivative of the function f(x) = 3x4−2x3+ 5x−7.
Solution:
To find the derivative of the function f(x) = 3x4−2x3+ 5x−7, we
need to differentiate each term with respect to x.
f′(x) = d
dx [3x4]−d
dx [2x3] + d
dx [5x]−d
dx [7]
Now, taking the derivatives term by term:
f′(x)=4·3x4−1−3·2x3−1+ 5 ·1−0
f′(x) = 12x3−6x2+ 5
Therefore, the derivative of the function f(x)=3x4−2x3+ 5x−7is
f′(x) = 12x3−6x2+ 5.
This is the solution to question 8.
Question 9
Question 9: Find the derivative of the function f(x)=3x4−2x3+
5x2−7x+ 1.
Solution:
Given function: f(x)=3x4−2x3+ 5x2−7x+ 1
To find the derivative of the function, we need to differentiate each
term separately using the power rule of differentiation.
9
The power rule states that if f(x) = xn, then f′(x) = nxn−1.
Now, let’s differentiate each term of the function:
f′(x) = d
dx (3x4)−d
dx (2x3) + d
dx (5x2)−d
dx (7x) + d
dx (1)
f′(x) = 12x3−6x2+ 10x−7+0
Therefore, the derivative of the function f(x)=3x4−2x3+5x2−7x+1
is f′(x) = 12x3−6x2+ 10x−7.
f′(x) = 12x3−6x2+ 10x−7Certainly! Here’s a question and its
step-by-step solution on Calculus for Liberty University in LateX
code.
Question 9: Find the derivative of the function f(x)=3x4−2x3+
5x2−7x+ 1.
Solution:
Given function: f(x)=3x4−2x3+ 5x2−7x+ 1
To find the derivative of the function, we need to differentiate each
term separately using the power rule of differentiation.
The power rule states that if f(x) = xn, then f′(x) = nxn−1.
Now, let’s differentiate each term of the function:
f′(x) = d
dx (3x4)−d
dx (2x3) + d
dx (5x2)−d
dx (7x) + d
dx (1)
f′(x) = 12x3−6x2+ 10x−7+0
Therefore, the derivative of the function f(x)=3x4−2x3+5x2−7x+1
is f′(x) = 12x3−6x2+ 10x−7.
f′(x) = 12x3−6x2+ 10x−7
Question 10
Question 10: Find the derivative of the function f(x)=4x3−2x2+
5x−1.
Step-by-step Solution: Given function: f(x) = 4x3−2x2+ 5x−1.
To find the derivative of the function, differentiate each term with
respect to xusing the power rule for differentiation:
f′(x) = d
dx (4x3)−d
dx (2x2) + d
dx (5x)−d
dx (1)
= 4 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
= 12x2−4x+ 5.
Therefore, the derivative of the function f(x)=4x3−2x2+ 5x−1is
f′(x) = 12x2−4x+ 5.
This is the required solution to question 10.Sure, here is a question
on Calculus along with its step-by-step solution in LateX code:
Question 10: Find the derivative of the function f(x)=4x3−2x2+
5x−1.
Step-by-step Solution: Given function: f(x) = 4x3−2x2+ 5x−1.
To find the derivative of the function, differentiate each term with
respect to xusing the power rule for differentiation:
10
f′(x) = d
dx (4x3)−d
dx (2x2) + d
dx (5x)−d
dx (1)
= 4 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
= 12x2−4x+ 5.
Therefore, the derivative of the function f(x)=4x3−2x2+ 5x−1is
f′(x) = 12x2−4x+ 5.
This is the required solution to question 10.
Question 11
Question 11: Consider the function f(x) = x3−3x2−9x+ 5. Find
the critical points and determine where the function is increasing or
decreasing.
Solution:
To find the critical points, we first need to find the derivative of
the function f(x).
Given: f(x) = x3−3x2−9x+ 5
The derivative of f(x)is found by differentiating each term sepa-
rately:
f′(x)=3x2−6x−9
Now, to find the critical points, we set the derivative equal to zero
and solve for x:
3x2−6x−9=0
x2−2x−3 = 0
We can factor the quadratic equation:
(x−3)(x+ 1) = 0
So, the critical points are x= 3 and x=−1.
To determine where the function is increasing or decreasing, we
analyze the sign of the derivative f′(x)in the intervals defined by the
critical points.
1. For x < −1: Pick a test point x=−2and substitute into f′(x):
f′(−2) = 3(−2)2−6(−2) −9 = 3(4) + 12 −9 = 15 >0
Therefore, the function is increasing on the interval x < −1.
2. For −1<x<3: Pick a test point x= 0 and substitute into f′(x):
f′(0) = 3(0)2−6(0) −9 = −9<0
Therefore, the function is decreasing on the interval −1<x<3.
11
3. For x > 3: Pick a test point x= 4 and substitute into f′(x):
f′(4) = 3(4)2−6(4) −9 = 3(16) −24 −9 = 27 >0
Therefore, the function is increasing on the interval x > 3.
Therefore, the function is increasing on (−∞,−1) ∪(3,∞)and de-
creasing on (−1,3).
This completes the solution.Sure, here is a Calculus question along
with the step-by-step solution in LateX code:
Question 11: Consider the function f(x) = x3−3x2−9x+ 5. Find
the critical points and determine where the function is increasing or
decreasing.
Solution:
To find the critical points, we first need to find the derivative of
the function f(x).
Given: f(x) = x3−3x2−9x+ 5
The derivative of f(x)is found by differentiating each term sepa-
rately:
f′(x) = 3x2−6x−9
Now, to find the critical points, we set the derivative equal to zero
and solve for x:
3x2−6x−9=0
x2−2x−3 = 0
We can factor the quadratic equation:
(x−3)(x+ 1) = 0
So, the critical points are x= 3 and x=−1.
To determine where the function is increasing or decreasing, we
analyze the sign of the derivative f′(x)in the intervals defined by the
critical points.
1. For x < −1: Pick a test point x=−2and substitute into f′(x):
f′(−2) = 3(−2)2−6(−2) −9 = 3(4) + 12 −9 = 15 >0
Therefore, the function is increasing on the interval x < −1.
2. For −1<x<3: Pick a test point x= 0 and substitute into f′(x):
f′(0) = 3(0)2−6(0) −9 = −9<0
Therefore, the function is decreasing on the interval −1<x<3.
3. For x > 3: Pick a test point x= 4 and substitute into f′(x):
f′(4) = 3(4)2−6(4) −9 = 3(16) −24 −9 = 27 >0
Therefore, the function is increasing on the interval x > 3.
Therefore, the function is increasing on (−∞,−1) ∪(3,∞)and de-
creasing on (−1,3).
This completes the solution.
12
Question 12
Find the antiderivative of the function f(x)=3x2−2x+ 5.
Step-by-step solution:
To find the antiderivative of the given function, we need to inte-
grate each term separately.
Given function: f(x)=3x2−2x+ 5
Integrating with respect to x:
Z3x2dx −Z2xdx +Z5dx
Integrating each term:
=3
3x3−2
2x2+ 5x+C
where Cis the constant of integration.
Simplifying:
=x3−x2+ 5x+C
Therefore, the antiderivative of the function f(x) = 3x2−2x+ 5 is
x3−x2+ 5x+C, where Cis the constant of integration.Question 12:
Find the antiderivative of the function f(x)=3x2−2x+ 5.
Step-by-step solution:
To find the antiderivative of the given function, we need to inte-
grate each term separately.
Given function: f(x)=3x2−2x+ 5
Integrating with respect to x:
Z3x2dx −Z2xdx +Z5dx
Integrating each term:
=3
3x3−2
2x2+ 5x+C
where Cis the constant of integration.
Simplifying:
=x3−x2+ 5x+C
Therefore, the antiderivative of the function f(x) = 3x2−2x+ 5 is
x3−x2+ 5x+C, where Cis the constant of integration.
13
Question 13
Question 13: Find the derivative of the function f(x) = x3−4x2+
5x−2.
Solution: To find the derivative of f(x) = x3−4x2+ 5x−2, we will
use the power rule for differentiation.
Given function, f(x) = x3−4x2+ 5x−2
Derivative of f(x)is given by:
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (5x)−d
dx (2)
Computing the derivatives:
f′(x) = 3x2−8x+ 5 −0
Therefore, the derivative of the function f(x) = x3−4x2+ 5x−2is
the function:
f′(x) = 3x2−8x+ 5
Absolutely! Here is a Calculus question along with step-by-step so-
lutions in LaTeX code:
Question 13: Find the derivative of the function f(x) = x3−4x2+
5x−2.
Solution: To find the derivative of f(x) = x3−4x2+ 5x−2, we will
use the power rule for differentiation.
Given function, f(x) = x3−4x2+ 5x−2
Derivative of f(x)is given by:
f′(x) = d
dx (x3)−d
dx (4x2) + d
dx (5x)−d
dx (2)
Computing the derivatives:
f′(x) = 3x2−8x+ 5 −0
Therefore, the derivative of the function f(x) = x3−4x2+ 5x−2is
the function:
f′(x) = 3x2−8x+ 5
Question 14
Question 14: Find the derivative of the following function:
f(x)=3x3−2x2+ 5x−7
Solution: To find the derivative of f(x) = 3x3−2x2+ 5x−7, we will
differentiate each term separately using the power rule.
14
f′(x) = d
dx 3x3−d
dx 2x2+d
dx (5x)−d
dx (7)
Applying the power rule:
f′(x) = 3 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying the derivatives:
f′(x) = 9x2−4x+ 5
Therefore, the derivative of the function f(x)=3x3−2x2+ 5x−7is
f′(x) = 9x2−4x+ 5.
This is question 14 for set 3.Certainly! Here is a question on
Calculus for Liberty University along with a step-by-step solution in
LaTeX code:
Question 14: Find the derivative of the following function:
f(x)=3x3−2x2+ 5x−7
Solution: To find the derivative of f(x) = 3x3−2x2+ 5x−7, we will
differentiate each term separately using the power rule.
f′(x) = d
dx 3x3−d
dx 2x2+d
dx (5x)−d
dx (7)
Applying the power rule:
f′(x) = 3 ·3x3−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying the derivatives:
f′(x) = 9x2−4x+ 5
Therefore, the derivative of the function f(x)=3x3−2x2+ 5x−7is
f′(x) = 9x2−4x+ 5.
This is question 14 for set 3.
Question 15
Question 15: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Solution: To find the derivative of the function f(x) = 3x2+ 5x+ 2,
we differentiate each term with respect to x. Using the power rule
for differentiation, we have:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
f′(x)=3·2x2−1+ 5 ·1x1−1+ 0
15
f′(x)=6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.Certainly! Here’s a Calculus question along with its
solution in LateX:
Question 15: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Solution: To find the derivative of the function f(x) = 3x2+ 5x+ 2,
we differentiate each term with respect to x. Using the power rule
for differentiation, we have:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
f′(x)=3·2x2−1+ 5 ·1x1−1+ 0
f′(x)=6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.
Question 16
Question 16: Find the derivative of the function f(x) = e3x2+2x.
Solution: To find the derivative of the given function f(x) = e3x2+2x,
we will use the chain rule.
Step 1: Identify the function and its components. Let u= 3x2+ 2x.
Then, our function can be written as f(x) = eu.
Step 2: Apply the chain rule. The chain rule states that if f(x) =
eu, then d
dx (eu) = eu·du
dx .
Step 3: Find du
dx .
du
dx =d
dx (3x2+ 2x) = 6x+ 2
Step 4: Substitute back into the chain rule formula.
d
dx (e3x2+2x) = e3x2+2x·(6x+ 2)
Therefore, the derivative of the function is f′(x) = e3x2+2x·(6x+
2).Sure, here is a Calculus question along with a step-by-step solution
in LateX format:
Question 16: Find the derivative of the function f(x) = e3x2+2x.
Solution: To find the derivative of the given function f(x) = e3x2+2x,
we will use the chain rule.
16
Step 1: Identify the function and its components. Let u= 3x2+ 2x.
Then, our function can be written as f(x) = eu.
Step 2: Apply the chain rule. The chain rule states that if f(x) =
eu, then d
dx (eu) = eu·du
dx .
Step 3: Find du
dx .
du
dx =d
dx (3x2+ 2x) = 6x+ 2
Step 4: Substitute back into the chain rule formula.
d
dx (e3x2+2x) = e3x2+2x·(6x+ 2)
Therefore, the derivative of the function is f′(x) = e3x2+2x·(6x+ 2).
Question 17
Question 17: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Step-by-step Solution: To find the derivative of the function f(x) =
3x2+ 5x+ 2, we use the power rule.
1. Differentiate each term of the function separately:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
2. Apply the power rule to each term:
f′(x) = 3 ·2x2−1+ 5 ·1x1−1+ 0
3. Simplify the derivatives:
f′(x) = 6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.Sure! Here is a Calculus question along with its step-
by-step solution in LateX code:
Question 17: Find the derivative of the function f(x)=3x2+ 5x+ 2.
Step-by-step Solution: To find the derivative of the function f(x) =
3x2+ 5x+ 2, we use the power rule.
1. Differentiate each term of the function separately:
f′(x) = d
dx (3x2) + d
dx (5x) + d
dx (2)
2. Apply the power rule to each term:
f′(x) = 3 ·2x2−1+ 5 ·1x1−1+ 0
3. Simplify the derivatives:
f′(x) = 6x+ 5
Therefore, the derivative of the function f(x) = 3x2+ 5x+ 2 is
f′(x) = 6x+ 5.
Question 18
Question 18: Find the derivative of the function f(x)=3x4+ 2x3−
5x2+ 7.
17
Solution: To find the derivative of the function f(x) = 3x4+ 2x3−
5x2+ 7, we will apply the power rule for derivatives to each term.
Given function: f(x)=3x4+ 2x3−5x2+ 7
Step 1: Find the derivative of 3x4
f′(x)=4·3x4−1= 12x3
Step 2: Find the derivative of 2x3
f′(x)=3·2x3−1= 6x2
Step 3: Find the derivative of −5x2
f′(x)=2·(−5)x2−1=−10x
Step 4: Find the derivative of the constant term 7
f′(x)=0
(The derivative of a constant term is 0)
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is
f′(x) = 12x3+ 6x2−10x
Feel free to reach out if you need further assistance or more ques-
tions!Certainly! Here is a Calculus question along with its step-by-
step solution in LateX code:
Question 18: Find the derivative of the function f(x)=3x4+ 2x3−
5x2+ 7.
Solution: To find the derivative of the function f(x) = 3x4+ 2x3−
5x2+ 7, we will apply the power rule for derivatives to each term.
Given function: f(x)=3x4+ 2x3−5x2+ 7
Step 1: Find the derivative of 3x4
f′(x)=4·3x4−1= 12x3
Step 2: Find the derivative of 2x3
f′(x)=3·2x3−1= 6x2
Step 3: Find the derivative of −5x2
f′(x) = 2 ·(−5)x2−1=−10x
Step 4: Find the derivative of the constant term 7
f′(x) = 0
(The derivative of a constant term is 0)
Therefore, the derivative of the function f(x) = 3x4+ 2x3−5x2+ 7
is
f′(x) = 12x3+ 6x2−10x
Feel free to reach out if you need further assistance or more ques-
tions!
18
Question 19
Question 19: Let f(x) = excos x. Find f′′ (x).
Solution: To find f′′ (x), we first need to find f′(x)and then take
its derivative.
Given: f(x) = excos x
Step 1: Find f′(x)
f′(x) = (excos x)′=ex(−sin x) + cos xex=ex(cos x−sin x)
Step 2: Find f′′ (x)
f′′ (x) = (ex(cos x−sin x))′=ex(−sin x−cos x)+ex(cos x−sin x) = ex(−sin x−cos x+cos x−sin x) = ex(−2 sin x) = −2exsin x
Therefore, the second derivative of f(x) = excos xis f′′ (x) = −2exsin x.Certainly!
Here is a Calculus question along with its step-by-step solution in La-
teX code:
Question 19: Let f(x) = excos x. Find f′′ (x).
Solution: To find f′′ (x), we first need to find f′(x)and then take
its derivative.
Given: f(x) = excos x
Step 1: Find f′(x)
f′(x) = (excos x)′=ex(−sin x) + cos xex=ex(cos x−sin x)
Step 2: Find f′′ (x)
f′′ (x)=(ex(cos x−sin x))′=ex(−sin x−cos x)+ex(cos x−sin x) = ex(−sin x−cos x+cos x−sin x) = ex(−2 sin x) = −2exsin x
Therefore, the second derivative of f(x) = excos xis f′′ (x) = −2exsin x.
Question 20
Step-by-step solution: To find the derivative of f(x)=2x3−5x2+
4x−7, we will differentiate each term with respect to xusing the
power rule.
1. Differentiate 2x3:
d
dx (2x3) = 2 ×3x3−1= 6x2
2. Differentiate −5x2:
d
dx (−5x2) = −5×2x2−1=−10x
3. Differentiate 4x:d
dx (4x)=4
19
4. Differentiate the constant term −7: The derivative of a constant
is always zero: d
dx (−7) = 0
Therefore, the derivative of f(x)=2x3−5x2+ 4x−7is:
f′(x)=6x2−10x+ 4
Question 20: Compute the derivative of the function f(x)=2x3−5x2+
4x−7.
Step-by-step solution: To find the derivative of f(x)=2x3−5x2+
4x−7, we will differentiate each term with respect to xusing the
power rule.
1. Differentiate 2x3:
d
dx (2x3) = 2 ×3x3−1= 6x2
2. Differentiate −5x2:
d
dx (−5x2) = −5×2x2−1=−10x
3. Differentiate 4x:d
dx (4x)=4
4. Differentiate the constant term −7: The derivative of a constant
is always zero: d
dx (−7) = 0
Therefore, the derivative of f(x)=2x3−5x2+ 4x−7is:
f′(x)=6x2−10x+ 4
Question 21
Find the derivative of the following function using the Power Rule:
f(x) = 3x5−2x3+ 4x−6
Solution:
To find the derivative of the function f(x)=3x5−2x3+ 4x−6using
the Power Rule, we differentiate each term separately.
1. Differentiate 3x5:
d
dx (3x5) = 5 ·3x5−1= 15x4
2. Differentiate −2x3:
d
dx (−2x3)=3·(−2)x3−1=−6x2
3. Differentiate 4x:d
dx (4x)=4
20
4. Differentiate −6(a constant):
d
dx (−6) = 0
Therefore, the derivative of the function f(x)=3x5−2x3+ 4x−6is:
f′(x) = 15x4−6x2+ 4
Question 21:
Find the derivative of the following function using the Power Rule:
f(x)=3x5−2x3+ 4x−6
Solution:
To find the derivative of the function f(x)=3x5−2x3+ 4x−6using
the Power Rule, we differentiate each term separately.
1. Differentiate 3x5:
d
dx (3x5) = 5 ·3x5−1= 15x4
2. Differentiate −2x3:
d
dx (−2x3)=3·(−2)x3−1=−6x2
3. Differentiate 4x:d
dx (4x)=4
4. Differentiate −6(a constant):
d
dx (−6) = 0
Therefore, the derivative of the function f(x)=3x5−2x3+ 4x−6is:
f′(x) = 15x4−6x2+ 4
Question 22
Question 22: Find the derivative of the function f(x)=3x4−2x2+5.
Solution: To find the derivative of the function f(x)=3x4−2x2+ 5,
we will differentiate each term with respect to xusing the power rule
of differentiation.
The power rule states that if the function is in the form f(x) = axn,
then the derivative is f′(x) = anxn−1.
Given f(x)=3x4−2x2+ 5, we differentiate each term as follows:
f′(x) = d
dx (3x4)−d
dx (2x2) + d
dx (5).
21
Applying the power rule: f′(x)=3·4x4−1−2·2x2−1+ 0.
Simplify the derivatives: f′(x) = 12x3−4x.
Therefore, the derivative of the function f(x) = 3x4−2x2+ 5 is
f′(x) = 12x3−4x.
I hope this helps! Let me know if you need any more assis-
tance.Certainly! Here’s a calculus question along with its solution
in LateX code:
Question 22: Find the derivative of the function f(x)=3x4−2x2+5.
Solution: To find the derivative of the function f(x)=3x4−2x2+ 5,
we will differentiate each term with respect to xusing the power rule
of differentiation.
The power rule states that if the function is in the form f(x) = axn,
then the derivative is f′(x) = anxn−1.
Given f(x)=3x4−2x2+ 5, we differentiate each term as follows:
f′(x) = d
dx (3x4)−d
dx (2x2) + d
dx (5).
Applying the power rule: f′(x)=3·4x4−1−2·2x2−1+ 0.
Simplify the derivatives: f′(x) = 12x3−4x.
Therefore, the derivative of the function f(x) = 3x4−2x2+ 5 is
f′(x) = 12x3−4x.
I hope this helps! Let me know if you need any more assistance.
Question 23
Question 23:
Find the derivative of the function f(x) = 3x2−2x+ 5.
Solution:
To find the derivative of the function f(x)=3x2−2x+5, we’ll apply
the power rule for differentiation.
Given function: f(x)=3x2−2x+ 5
Step 1: Find the derivative of each term separately.
d
dx (3x2)=2·3x2−1= 6x
d
dx (−2x) = −2
Step 2: Combine the derivatives of individual terms.
The derivative of the function f(x)=3x2−2x+ 5 is:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
f′(x)=6x−2+0
f′(x)=6x−2
22
Therefore, the derivative of the function f(x)=3x2−2x+5 is f′(x) =
6x−2.Sure, here is a question on Calculus for Liberty University along
with its step-by-step solution presented in LateX code:
Question 23:
Find the derivative of the function f(x) = 3x2−2x+ 5.
Solution:
To find the derivative of the function f(x)=3x2−2x+5, we’ll apply
the power rule for differentiation.
Given function: f(x)=3x2−2x+ 5
Step 1: Find the derivative of each term separately.
d
dx (3x2)=2·3x2−1= 6x
d
dx (−2x) = −2
Step 2: Combine the derivatives of individual terms.
The derivative of the function f(x)=3x2−2x+ 5 is:
f′(x) = d
dx (3x2)−d
dx (2x) + d
dx (5)
f′(x)=6x−2+0
f′(x)=6x−2
Therefore, the derivative of the function f(x) = 3x2−2x+ 5 is
f′(x) = 6x−2.
Question 24
Question 24: Find the derivative of the function f(x)=3x4−8x2+
5x−2.
Solution: To find the derivative of f(x), we can apply the power
rule for differentiation. The power rule states that if f(x) = axn, then
f′(x) = naxn−1.
Given function: f(x)=3x4−8x2+ 5x−2
Applying the power rule:
f′(x) = 4(3)x4−1−2(8)x2−1+ 5x0
Simplifying the derivative:
f′(x) = 12x3−16x+ 5
Therefore, the derivative of the function f(x)=3x4−8x2+ 5x−2is
f′(x) = 12x3−16x+ 5.Sure, here is a calculus question along with its
step-by-step solution in LateX code:
23
Question 24: Find the derivative of the function f(x)=3x4−8x2+
5x−2.
Solution: To find the derivative of f(x), we can apply the power
rule for differentiation. The power rule states that if f(x) = axn, then
f′(x) = naxn−1.
Given function: f(x)=3x4−8x2+ 5x−2
Applying the power rule:
f′(x) = 4(3)x4−1−2(8)x2−1+ 5x0
Simplifying the derivative:
f′(x) = 12x3−16x+ 5
Therefore, the derivative of the function f(x)=3x4−8x2+ 5x−2is
f′(x) = 12x3−16x+ 5.
Question 25
Step 1: Determine the critical points by finding the derivative of
the function:
f′(x)=3x2−6x
Step 2: Set f′(x)=0to find the critical points:
3x2−6x= 0
3x(x−2) = 0
So, the critical points are x= 0 and x= 2.
Step 3: Find the function values at the critical points and end-
points of the interval: - f(0) = 03−3(0)2+ 2 = 2 -f(3) = 33−3(3)2+2 = 2
-f(2) = 23−3(2)2+ 2 = −2
Step 4: Compare the function values at critical points and end-
points to determine the absolute maximum and minimum: - The
maximum value is f(0) = 2 at x= 0 - The minimum value is f(2) = −2
at x= 2
Therefore, the absolute maximum value of the function on the
interval [0,3] is 2 and the absolute minimum value is -2.Question 25:
Find the absolute maximum and minimum values of the function
f(x) = x3−3x2+ 2 on the interval [0,3].
Step 1: Determine the critical points by finding the derivative of
the function:
f′(x)=3x2−6x
Step 2: Set f′(x)=0to find the critical points:
3x2−6x= 0
24
3x(x−2) = 0
So, the critical points are x= 0 and x= 2.
Step 3: Find the function values at the critical points and end-
points of the interval: - f(0) = 03−3(0)2+ 2 = 2 -f(3) = 33−3(3)2+2 = 2
-f(2) = 23−3(2)2+ 2 = −2
Step 4: Compare the function values at critical points and end-
points to determine the absolute maximum and minimum: - The
maximum value is f(0) = 2 at x= 0 - The minimum value is f(2) = −2
at x= 2
Therefore, the absolute maximum value of the function on the
interval [0,3] is 2 and the absolute minimum value is -2.
Question 26
Question 26: Find the derivative of the function f(x)=3x4−2x2+
5x−7.
Step-by-step Solution: To find the derivative of the function f(x) =
3x4−2x2+5x−7, we will use the power rule of differentiation. Applying
the power rule to each term of the function, we get:
f′(x) = 4 ·3x4−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying, we have:
f′(x) = 12x3−4x+ 5
Therefore, the derivative of the function f(x)=3x4−2x2+ 5x−7is
f′(x) = 12x3−4x+ 5.Sure, here is a question on calculus along with its
step-by-step solutions in LateX code:
Question 26: Find the derivative of the function f(x)=3x4−2x2+
5x−7.
Step-by-step Solution: To find the derivative of the function f(x) =
3x4−2x2+5x−7, we will use the power rule of differentiation. Applying
the power rule to each term of the function, we get:
f′(x) = 4 ·3x4−1−2·2x2−1+ 5 ·1x1−1−0
Simplifying, we have:
f′(x) = 12x3−4x+ 5
Therefore, the derivative of the function f(x)=3x4−2x2+ 5x−7is
f′(x) = 12x3−4x+ 5.
25
Question 27
Step 1: Apply the Power Rule to each term of the function.
f′(x) = d
dx (3x4)−d
dx (8x2) + d
dx (5)
= 3 ·4x4−1−8·2x2−1+ 0
= 12x3−16x
Step 2: Simplify the derivative.
f′(x) = 12x3−16x
Therefore, the derivative of the function f(x) = 3x4−8x2+ 5 is
f′(x) = 12x3−16x.Question 27: Find the derivative of the following
function using the Power Rule.
f(x)=3x4−8x2+ 5
Step 1: Apply the Power Rule to each term of the function.
f′(x) = d
dx (3x4)−d
dx (8x2) + d
dx (5)
= 3 ·4x4−1−8·2x2−1+ 0
= 12x3−16x
Step 2: Simplify the derivative.
f′(x) = 12x3−16x
Therefore, the derivative of the function f(x) = 3x4−8x2+ 5 is
f′(x) = 12x3−16x.
Question 28
Question 28: Find the derivative of the function f(x)=3x2+ 4x−5.
Step-by-step Solution: To find the derivative of the function, we
apply the power rule for derivatives. The power rule states that the
derivative of xnis nxn−1.
Given function: f(x)=3x2+ 4x−5
1. Find the derivative of 3x2:
d
dx (3x2)=2·3x2−1= 6x
2. Find the derivative of 4x:
d
dx (4x)=4·x1−1= 4
26
3. Find the derivative of the constant term −5: The derivative of
a constant is 0.
Therefore, the derivative of the function f(x)=3x2+ 4x−5is:
f′(x)=6x+ 4
Certainly! Here is a Calculus question along with the step-by-step
solution presented in LaTeX code:
Question 28: Find the derivative of the function f(x)=3x2+ 4x−5.
Step-by-step Solution: To find the derivative of the function, we
apply the power rule for derivatives. The power rule states that the
derivative of xnis nxn−1.
Given function: f(x)=3x2+ 4x−5
1. Find the derivative of 3x2:
d
dx (3x2)=2·3x2−1= 6x
2. Find the derivative of 4x:
d
dx (4x)=4·x1−1= 4
3. Find the derivative of the constant term −5: The derivative of
a constant is 0.
Therefore, the derivative of the function f(x)=3x2+ 4x−5is:
f′(x)=6x+ 4
Question 29
Question 29: Find the derivative of the function f(x)=5x3−2x2+
4x+ 7.
Step-by-step solution: To find the derivative of the function f(x) =
5x3−2x2+ 4x+ 7, we will differentiate each term using the power rule.
f′(x) = d
dx (5x3)−d
dx (2x2) + d
dx (4x) + d
dx (7)
= 5 ·3x3−1−2·2x2−1+ 4 ·1x1−1+ 0
= 15x2−4x+ 4.
Therefore, the derivative of the function f(x)=5x3−2x2+ 4x+ 7
is f′(x) = 15x2−4x+ 4.Certainly! Here is a question on Calculus
for Liberty University along with the step-by-step solution in LateX
code:
Question 29: Find the derivative of the function f(x)=5x3−2x2+
4x+ 7.
27
Step-by-step solution: To find the derivative of the function f(x) =
5x3−2x2+ 4x+ 7, we will differentiate each term using the power rule.
f′(x) = d
dx (5x3)−d
dx (2x2) + d
dx (4x) + d
dx (7)
= 5 ·3x3−1−2·2x2−1+ 4 ·1x1−1+ 0
= 15x2−4x+ 4.
Therefore, the derivative of the function f(x)=5x3−2x2+ 4x+ 7 is
f′(x) = 15x2−4x+ 4.
Question 30
Question: Calculate the derivative of the function f(x) = x3−2x2+
5x−3.
Step-by-step Solution: To find the derivative of the function f(x) =
x3−2x2+ 5x−3, we will differentiate each term separately using the
power rule for derivatives.
1. Find the derivative of the term x3:
d
dx (x3) = 3x3−1= 3x2
2. Find the derivative of the term −2x2:
d
dx (−2x2) = −2·2x2−1=−4x
3. Find the derivative of the term 5x:
d
dx (5x) = 5 ·1=5
4. Find the derivative of the constant term −3:
d
dx (−3) = 0
Therefore, the derivative of the function f(x) = x3−2x2+ 5x−3is:
f′(x) = 3x2−4x+ 5
This completes the solution.
I hope this helps! Let me know if you need any further assis-
tance.Certainly! Here is a Calculus question along with its step-by-
step solution in LateX code:
Question: Calculate the derivative of the function f(x) = x3−2x2+
5x−3.
28
Step-by-step Solution: To find the derivative of the function f(x) =
x3−2x2+ 5x−3, we will differentiate each term separately using the
power rule for derivatives.
1. Find the derivative of the term x3:
d
dx (x3) = 3x3−1= 3x2
2. Find the derivative of the term −2x2:
d
dx (−2x2) = −2·2x2−1=−4x
3. Find the derivative of the term 5x:
d
dx (5x) = 5 ·1=5
4. Find the derivative of the constant term −3:
d
dx (−3) = 0
Therefore, the derivative of the function f(x) = x3−2x2+ 5x−3is:
f′(x) = 3x2−4x+ 5
This completes the solution.
I hope this helps! Let me know if you need any further assistance.
29
