MATH225 – Calculus Week 3 Test-Grade 85% 2025-2026 American Public University

MATH225 – Calculus Week 3 Test-Grade 85% 2025-2026 American Public University

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Questions 20 Questions

Limits 180 Minutes

Attempt 1

Points

Due Date

100 pts possible

No due date.

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)=(3x 2+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

f′(x)=3(3x2+7x)4(2x3−6x)2(6x2−6)(2x3−6x)6−4(2x3−6x)3(3x2+7x)3(6x+7)(2x3−6x)6f′(x)=3(3x2+7x)4(2x3−6x)2(6x2−6) (2x3−6x)6−4(2x3−6x)3(3x2+7x)3(6x+7)(2x3−6x)6

To find the derivative, use the quotient rule, applying the chain rule as needed. f′=(gh)′=h⋅g′− g⋅h′h2g(x)=(3x2+7x)4g′(x)=4(3x2+7x)3(6x+7)h(x)=(2x3−6x)3h′(x)=3(2x3−6x)2(6x2−6)f′=(gh)′=h⋅g′− g⋅h′h2g(x)=(3x2+7x)4g′(x)=4(3x2+7x)3(6x+7)h(x)=(2x3−6x)3h′(x)=3(2x3−6x)2(6x2−6) Therefore, 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

5/5 Question 5 : 5 ptsSkip to question text.

Find f′(x) if f(x)=cos(2x2).Find f′(x) if f(x)=cos(2x2). f′(x)=−2xsin(2x2)f′(x)=−2xsin(2x2) f′(x)=4xsin(2x2)f′(x)=4xsin(2x2) f′(x)=−4xsin(2x2)f′(x)=−4xsin(2x2) f′(x)=−sin(4x)f′(x)=−sin(4x) You can find f′(x) using the chain rule.f(x)=cos(2x2)f′(x)=−sin(2x2)⋅ddx(2x2)=−sin(2x2)⋅4x=−4xsin(2x2)You can find f′ (x) using the chain rule.f(x)=cos(2x2)f′(x)=−sin(2x2)⋅ddx(2x2)=−sin(2x2)⋅4x=−4xsin(2x2) 5/5 Question 6 : 5 ptsSkip to question text.

Find the derivative of the functionf(x)=2x.Find the derivative of the functionf(x)=2x. f′(x)=2x⋅2f′(x)=2x⋅2 f′(x)=2xf′(x)=2x f′(x)=x⋅2x−1f′(x)=x⋅2x−1 f′(x)=2x⋅ln2f′(x)=2x⋅ln2 Use the rule for the derivative of an exponential function f(x)=N x, 'where N>0.f′(x)=NxlnNIf f(x)=2x, then f′ (x)=2xln2.Use the rule for the derivative of an exponential function f(x)=Nx, 'where N>0.f′(x)=NxlnNIf f(x)=2x, then f′ (x)=2xln2.

5/5 Question 7 : 5 ptsSkip to question text.

Find the derivative. f(x)=cosx2f(x)=cosx2 f′(x)=−sinxf′(x)=−sinx f′(x)=−12cosxf′(x)=−12cosx f′(x)=12sinxf′(x)=12sinx

f′(x)=−12sinxf′(x)=−12sinx None of these

Using the constant multiple rule, movethe coefficient outside of the functionbefore differentiating:f(x)=12(cosx).So, f' (x)=12(cosx)'The derivative of cosx is −sinx,So, f'(x)=12(−sinx)=−12sinx. Using the constant multiple rule, movethe co efficient outside of the functionbefore differentiating:f(x)=12(cosx).So, f'(x)=12(cosx)'The derivative of cosx is −sinx,So, f'( x)=12(−sinx)=−12sinx.

5/5

Question 8 : 5 ptsSkip to question text.

Given y=xsin(1x), find y'.Given y=xsin(1x), find y'. y′=sin(1x)−1xcos(1x)y′=sin(1x)−1xcos(1x) y′=sin(1x)+1xcos(1x)y′=sin(1x)+1xcos(1x) y′=sin(1x)+1x2cos(1x)y′=sin(1x)+1x2cos(1x) y′=1x2sin(1x)−cos(1x)y′=1x2sin(1x)−cos(1x) Use the product rule and the chainrule to compute this derivative.y=xsin(1x)y'=x⋅ddxsin(1x) +sin(1x)⋅ddxxy'=x⋅cos(1x)⋅ddx(1x)+sin(1x)y'=xcos(1x)⋅(−1)x −2+sin(1x)y'=sin(1x)−(1x)cos(1x)Use the product rule and the chainrule to compute this derivative.y=xsin(1x)y'=x⋅ddxsin(1x)+sin(1x)⋅ddxxy'=x⋅cos(1x)⋅ddx(1x) +sin(1x)y'=xcos(1x)⋅(−1)x−2+sin(1x)y'=sin(1x)−(1x)cos(1x) 5/5 Question 9 : 5 ptsFind the derivative. f (x) = sin x cos x

Find the derivative. f (x) = sin x cos x f′(x)=−cosxsinxf′(x)=−cosxsinx f′(x)=(sinxcosx)−1f′(x)=(sinxcosx)−1 f′(x)=−sin2x−cos2xcos2xf′(x)=−sin2x−cos2xcos2x f′(x)=−sin2x+cos2xf′(x)=−sin2x+cos2x None of these

f(x)=sinxcosxf′(x)=−sin2x+cos2xWrite the function.Apply the product rule.f(x)=sinxcosxWrite the function.f′ (x)=−sin2x+cos2xApply the product rule.

5/5

Question 10 : 5 ptsSkip to question text.

Find dydx, wherey=log2(x2+1).Find dydx, wherey=log2(x2+1). dydx=1ln2⋅(x2+1)dydx=1ln2⋅(x2+1) dydx=2x(x2+1)dydx=2x(x2+1) dydx=2xln2⋅(x2+1)dydx=2xln2⋅(x2+1) dydx=2xln(x2−1)dydx=2xln(x2−1)

We can rewrite the log expression in terms of the natural log using thechange of base formula.y=log2(x2+1)=ln(x2+1 )ln2Take the derivative.dydx=1ln2⋅1x2+1⋅ddx(x2+1)=1ln2⋅1x2+1⋅2x=2xln2⋅(x2+1)We can rewrite the log expression in terms of the natural log using thechange of base formula.y=log2(x2+1)=ln(x2+1)ln2Take the derivative.dydx=1ln2⋅1x2+1 ⋅ddx(x2+1)=1ln2⋅1x2+1⋅2x=2xln2⋅(x2+1) 5/5 Question 11 : 5 ptsFind the derivative. f (t) = cos2t

Find the derivative. f (t) = cos2t f ′(t) = 2 cos t sin t f ′(t) = 2 cos t f ′(t) = −2 sin t f ′(t) = −2 cos t sin t None of these

f(t)=cos2tf′(t)=2cost⋅ddt(cost)=2cost(−sint)=−2costsintWrite the function.Apply the power rule.The derivativeof cos t is −sin t.Simplify.f(t)=cos2tWrite the function.f′(t)=2cost⋅ddt(cost)Apply the power rule.=2cost(−sint)The derivativeof cos t is −sin t.=−2costsintSimplify.

5/5

Question 12 : 5 ptsSkip to question text.

Find the derivative of f(x)=x−−√4.Find the derivative of f(x)=x4. f′(x)=12x−−√f′(x)=12x f′(x)=14x−−√f′(x)=14x f′(x)=18x−−√f′(x)=18x f′(x)=x−−√8f′(x)=x8 None of these

Compute f′(x).f′(x)=lim△x→0f(x+△x)−f(x)△x= lim △x→0 x +△x−−−−−−√4−x−−√4△x=lim △x→0x +△x−−−−−−√−x− −√4△x=lim△x→0x+△x−x4△x(x +△x−−−−−−√+x−−√)=lim△x→014(x+△x−−−−−−√+x−−√)=18x−−√Use the definition of the derivative.Substitute.Simplify.Multiply the numerator anddenominator byx +△x−−−−−−√+x− −√.Simplify.Evaluate the limit.Compute f′(x).f′(x)=lim△x→0f(x+△x)−f(x)△xUse the definition of the derivative.= lim△x→0 x +△x4−x4△xSubstitute.=lim△x→0x +△x−x4△xSimplify.=lim△x→0x+△x−x4△x(x +△x+x)Multiply the numerator anddenomin ator byx +△x+x.=lim△x→014(x+△x+x)Simplify.=18xEvaluate the limit.

5/5

Question 13 : 5 ptsSkip to question text.

Find the derivative of P(t).P(t)=−(2t 2−14t+4)2Find the derivative of P(t).P(t)=−(2t2−14t+4)2 P′(t)=2(2t2−14t+4)(4t−14)P′(t)=2(2t2−14t+4)(4t−14) P′(t)=−2(2t2−14t+4)(4t−14)P′(t)=−2(2t2−14t+4)(4t−14)

P′(t)=−2(4t−14)P′(t)=−2(4t−14) P′(t)=2(2t2−14t+4)(4t−14)−1P′(t)=2(2t2−14t+4)(4t−14)−1 To find the derivative, first take the derivativeof the exponent.−2(2t 2−14t+4)Then multiply that piece by the derivati ve ofwhat is inside the parentheses.−2(2t2−14t+4)(4t−14)To find the derivative, first take the derivativeof the exponent .−2(2t2−14t+4)Then multiply that piece by the derivative ofwhat is inside the parentheses.−2(2t2−14t+4)(4t−14) 5/5 Question 14 : 5 ptsSkip to question text.

Find dydx given thaty=3u2, u=2v, and v=x6.Find dydx given thaty=3u2, u=2v, and v=x6. dydx=2x3dydx=2x3 dydx=6vdydx=6v dydx=13dydx=13 dydx=x3dydx=x3 To find the derivative, use the chain rule twice.dydx=dydu⋅dudv⋅dvdxSince y=3u 2, dydu=6u.Since u=2v, dudv=2.Sin ce v=x6, dvdx=16.Substitute to find dydx.dydx=dydu⋅dudv⋅dvdx=6u⋅2⋅16=2uNow write 2u in terms of x.dydx=2u=2( 2v)=4v=4(x6)=2x3To find the derivative, use the chain rule twice.dydx=dydu⋅dudv⋅dvdxSince y=3u2, dydu=6u.Since u= 2v, dudv=2.Since v=x6, dvdx=16.Substitute to find dydx.dydx=dydu⋅dudv⋅dvdx=6u⋅2⋅16=2uNow write 2u in terms of x.dy dx=2u=2(2v)=4v=4(x6)=2x3 5/5 Question 15 : 5 ptsSkip to question text.

Find the derivative of f(x)=23x.Find the derivative of f(x)=23x. f′(x)=−32x2f′(x)=−32x2 f′(x)=−23x2f′(x)=−23x2 f′(x)=23x2f′(x)=23x2 f′(x)=32x2f′(x)=32x2 f(x)=23xf′(x)=limΔx→023(x+Δx) −23xΔx=limΔx→02x−2(x+Δx)3x(x+Δx)Δx=limΔx→02x−2x−2Δx3x(x+Δx)Δx=limΔx→0−2Δx3x(x+Δx)Δx=limΔx→0−23x(x+Δ x)=−23x2 f(x)=23xf′(x)=limΔx→023(x+Δx) −23xΔx=limΔx→02x−2(x+Δx)3x(x+Δx)Δx=limΔx→02x−2x−2Δx3x(x+Δx)Δx=limΔx→0−2Δx3x(x+Δx)Δx=limΔx→0−23x(x+ Δx)=−23x2 5/5 Question 16 : 5 ptsSkip to question text.

Find the derivative of f(x)=−15–√x.Find the derivative of f(x)=−15x. f′(x)=15–√x2f′(x)=15x2 f′(x)=−15–√x2f′(x)=−15x2 f′(x)=5–√x2f′(x)=5x2

f′(x)=−5–√x2f′(x)=−5x2 f(x)=−15–√xf′(x)=limΔx→0−15–√(x+Δx)−−15–√xΔx=limΔx→0−x+(x+Δx)5–√x(x+Δx)Δx=limΔx→0−x+x+Δx5– √x(x+Δx)Δx=limΔx→0Δx5–√x(x+Δx)Δx=limΔx→015–√x(x+Δx)=15–√x2 f(x)=−15xf′(x)=limΔx→0−15(x+Δx)− −15xΔx=limΔx→0−x+ (x+Δx)5x(x+Δx)Δx=limΔx→0−x+x+Δx5x(x+Δx)Δx=limΔx→0Δx5x(x+Δx)Δx=limΔx→015x(x+Δx)=15x2 0 /5 Question 17 : 5 ptsSkip to question text.

If h=3k2 and k=(3t)2, then find dhdt.If h=3k2 and k=(3t)2, then find dhdt. dhdt=972t3dhdt=972t3 dhdt=486t3dhdt=486t3 dhdt=324t3dhdt=324t3 dhdt=108t3dhdt=108t3 To find the derivative, use the chain rule.dhdt=dhdk⋅dkdtSince h=3k 2, dhdk=6k.Since k=(3t)2, dkdt=2(3t) (3)=18t.Substitute to find dhdt.dhdt=dhdk⋅dkdt=6k⋅18t=108ktNow subsitute (3t) 2 for k.dhdt=108(3t)2t=108⋅9t2⋅t=9 72t3To find the derivative, use the chain rule.dhdt=dhdk⋅dkdtSince h=3k2, dhdk=6k.Since k=(3t)2, dkdt=2(3t) (3)=18t.Substitute to find dhdt.dhdt=dhdk⋅dkdt=6k⋅18t=108ktNow subsitute (3t)2 for k.dhdt=108(3t)2t=108⋅9t2⋅t=972t3 5 /5 Question 18 : 5 ptsSkip to question text.

Suppose f(x)=2x−1−−−−−√. What is the slope of the line tangent to f when x=5?Suppose f(x)=2x−1. What is the slope of the l ine tangent to f when x=5? m=16–√m=16 m=111−−√m=111 m=12m=12 m=13m=13 None of these

First compute f'(5). This can be done either by first computing f'(x) and then substituting 5 for x in the resulting expression or by computing f'(5).

Use the definition of the derivative to compute f'(5). f′(5)=lim△x→0f(5+△x)−f(5)△xf′(5)=lim△x→0f(5+△x)−f(5)△x Substitute. = lim△x→0 2(5 +△x)−1−−−−−−−−−−−−√−2(5)−1−−−−−−−√△x = lim△x→0 2(5 +△x)−1−2(5)−1△x Simplify. =lim△x→09+2△x−−−−−−−−√−9–√△x=lim△x→09+2△x−9△x Multiply the numerator and denominator by 9+2△x−−−−−−−−√+9–√.Multiply the numerator and denominator by 9+ 2△x+9. =lim△x→09+2△x−9△x(9+2△x−−−−−−−−√+9–√)=lim△x→09+2△x−9△x(9+2△x+9) Simplify.

Evaluate the limit.

=lim△x→029+2△x−−−−−−−−√+9–√=lim△x→029+2△x+9 =229–√=229

Simplify. =13=13 The slope of the tangent line at x=5 is m=13.The slope of the tangent line at x=5 is m=13.

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Question 19 : 5 ptsIf f (x) = x 3 + 3x, and f −1 (4) = 1, find the derivative of f −1 at x = 4.

If f (x) = x 3 + 3x, and f −1 (4) = 1, find the derivative of f −1 at x = 4.

1/5 2x 2 + 3 1/51 1/6

If f(x)=x3+3x, then f′(x)=3x2+3>0.Hence f is monotonically increasing and thusinvertible. If f(x)=x3+3x, then f′ (x)=3x2+3>0.Hence f is monotonically increasing and thusinvertible. ddx[f−1(x)]=1f′(f−1(x))=13(f−1(x))2+3At x=4, f−1(4)=1, therefore,(f−1)′(4)=13∙1+3=16ddx[f−1(x)]=1f′ (f−1(x))=13(f−1(x))2+3At x=4, f−1(4)=1, therefore,(f−1)′(4)=13•1+3=16 0/5

Question 20 : 5 ptsIf f (x) = sin x + ex + x and f −1 (1) = 0, find the derivative of f −1 at x = 1.

If f (x) = sin x + e x + x and f −1 (1) = 0, find the derivative of f −1 at x = 1.

1cos 1+e+11cos 1+e+1 1/3 1/2 12+e12+e

If f(x)=sinx+ex+x, thenf′(x)=cosx+ex+1>0, becausethe exponential function is alwayspositive and cosx≥−1. Hence f i smonotonically increasing and thusinvertible. If f(x)=sinx+ex+x, thenf′(x)=cosx+ex+1>0, becausethe exponential functi on is alwayspositive and cosx≥−1. Hence f ismonotonically increasing and thusinvertible. ddx[f−1(x)]=1f′(f−1(x))=1cos(f−1(x))+ef−1(x)+1at x=1, f−1(1)=0; therefore,(f−1)′(1)=1cos(f−1(1)) +ef−1(1)+1=1cos(0)+e0+1=13ddx[f−1(x)]=1f′(f−1(x))=1cos(f−1(x))+ef−1(x)+1at x=1, f−1(1)=0; therefore, (f−1)′(1)=1cos(f−1(1))+ef−1(1)+1=1cos(0)+e0+1=13 5/5

Quiz Score: 85 out of 100