MATH 225 Week 3 Test SUMMER 2025

Week 3 Test - Grade: 85%
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Questions Limits Points Due Date


20 Questions 180 Minutes 100 pts possible No due date.




Attempt 1 85% (85 of 100) Completed on 03/23/25 at 07:18PM
Score for this quiz: 85% ( 85 /100)
Submitted Mar 23 at 7:18pm
This attempt took 35 minutes.

Question 1 : 5 ptsSuppose f (x) = ln x2. Find f '(x).
Suppose f (x) = ln x2. Find f '(x).
f ' (x) = 2xlnx2
f′(x)=1xf′(x)=1x f′
(x)=2xf′(x)=2x f′
(x)=1x2f′(x)=1x2
Use a property of logarithms torewrite the function. Then,find the derivative.f(x)=lnx2=2lnxf′(x)=2xUse a property of l
ogarithms torewrite the function. Then,find the derivative.f(x)=lnx2=2lnxf′(x)=2x
5/5


Question 2 : 5 ptsSkip to question text.

, Find the derivative of:f(x)=x3(3x2)2Find the derivative of:f(x)=x3(3x2)2 f′
(x)=−19x2f′(x)=−19x2 f′(x)=(3x2)⋅3x2−x3⋅6x(3x2)2f′
(x)=(3x2)⋅3x2−x3⋅6x(3x2)2 f′(x)=−13x2f′(x)=−13x2 f′
(x)=(3x2)2⋅3x2+x3⋅2(3x2)(6x)(3x2)4f′(x)=(3x2)2⋅3x2+x3⋅2(3x2)(6x)(3x2)4
To find the derivative, use the quotient rule.f′(x)=(3x2)2⋅3x2−x3⋅2(3x2)(6x)(3x2)4This equation may be simplified tof′
(x)=27x6−36x634x8=32x6(3−4)32x6⋅32x2.The reduced answer is much simpler.f′(x)=−19x2Notice that you could have s
implified andreduced this function for your first step.Then the solution could be found usingonly the power rule.To fin
d the derivative, use the quotient rule.f′(x)=(3x2)2 ⋅3x2−x3 ⋅2(3x2)(6x)(3x2)4This equation may be simplified tof′
(x)=27x6−36x634x8=32x6(3−4)32x6⋅32x2.The reduced answer is much simpler.f′(x)=−19x2Notice that you could have si
mplified andreduced this function for your first step.Then the solution could be found usingonly the power rule.
5/5


Question 3 : 5 ptsSkip to question text.
Find the derivative.f(x)=3(4x+7)4−4(3x+7)3Find the derivative.f(x)=3(4x+7)4−4(3x+7)3
f′(x)=12(4x+7)3−12(3x+7)2f′(x)=12(4x+7)3−12(3x+7)2 f′(x)=48(4x+7)3⋅36(3x+7)2f′
(x)=48(4x+7)3⋅36(3x+7)2 f′(x)=12(4x+7)3⋅12(3x+7)2f′(x)=12(4x+7)3⋅12(3x+7)2 f′
(x)=48(4x+7)3−36(3x+7)2f′(x)=48(4x+7)3−36(3x+7)2
To find the derivative, consider each piece separated by a plus or minus sign as its own problem.

So 3(4x + 7)4 is one problem and −4(3x + 7)3 is the other.

To solve the first piece, multiply the 3 by the exponent (which is 4) and then reduce the exponent by 1. Then multiply all of that by the derivative of
what is inside the parentheses. This gives you 48(4x + 7)3.

To solve the other piece, multiply the −4 by the exponent (which is 3) and then reduce the exponent by 1. Then multiply all of that by the derivative
of what is inside the parentheses. This gives you −36(3x + 7)2.
0/5


Question 4 : 5 ptsSkip to question text.
Find the derivative.f(x)=(3x2+7x)4(2x3−6x)3Find the derivative.f(x)=(3x2+7x)4(2x3−6x)3 f′(x)=4(2x3−6x)3(3x2+7x)3(6x+7)
(2x3−6x)6+3(3x2+7x)4(2x3−6x)2(6x2−6)(2x3−6x)6f′(x)=4(2x3−6x)3(3x2+7x)3(6x+7) (2x3−6x)6+3(3x2+7x)4(2x3−6x)2(6x2−6)
(2x3−6x)6 f′(x)=4(2x3−6x)3(3x2+7x)3(6x+7)−3(3x2+7x)4(2x3−6x)2(6x2−6)f′
(x)=4(2x3−6x)3(3x2+7x)3(6x+7)−3(3x2+7x)4(2x3−6x)2(6x2−6) f′(x)=4(2x3−6x)3(3x2+7x)3(6x+7)
(2x3−6x)6−3(3x2+7x)4(2x3−6x)2(6x2−6)(2x3−6x)6f′(x)=4(2x3−6x)3(3x2+7x)3(6x+7) (2x3−6x)6−3(3x2+7x)4(2x3−6x)2(6x2−6)
(2x3−6x)6