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AP Calculus AB : Computation of the Derivative

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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Example Question #21 : Chain Rule And Implicit Differentiation

Find the derivative of:

$k(t)=\sin(\cos t)$

Possible Answers:

$k'(t)=\cos(\cos t)\cdot(-\sin t)$

$k'(t)=\cos(\cos t)$

$k'(t)=-\cos(\cos t)\cdot(-\sin t)$

$k'(t)=\sec(\cos t)\cdot(-\sin t)$

$k'(t)=\cos(\cos t)\cdot \sin t$

Correct answer:

$k'(t)=\cos(\cos t)\cdot(-\sin t)$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

$g(t)=\sin t$

and

$f(t)=\cos t$ .

Since

$g'(t)=\cos t$

and

$f'(t)=-\sin t$ ,

we can conclude that

$k'(t)=\frac{d}{dt}g(f(t))=\cos(f(t))\cdot(-\sin t)$

and

$k'(t)=\cos(\cos t)\cdot(-\sin t)$

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Example Question #28 : Chain Rule And Implicit Differentiation

Find the derivative of:

$f(x)=\cos(1+\tan2x)$

Possible Answers:

$f'(x)=-\sin(1+\tan 2x)$

$f'(x)=-\sin(1+\tan 2x)\sec^2(2x)$

$f'(x)=-2\sin(1+\tan 2x)\sec^2(2x)$

$f'(x)=-2\sin(1+\tan 2x)(1+\sec^2(2x))$

$f'(x)=-2\sin(1+\tan 2x)\sec(2x)\tan(2x)$

Correct answer:

$f'(x)=-2\sin(1+\tan 2x)\sec^2(2x)$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dx}g(h(j(x)))=g'(h(j(x)))h'(j(x))j'(x)$

So in this problem we let

$g(x)=\cos x$

and

$h(x)=1+\tan x$

and

j(x)=2x .

Since

$g'(x)=-\sin x$

and

$h'(x)=\sec^2(x)$

and

j'(x)=2 ,

we can conclude that

$f'(x)=\frac{d}{dx}g(h(j(x)))=-sin(h(j(x)))\cdot (sec^2(j(x)))\cdot (-2)$

$f'(x)=-2\sin(1+\tan2x)(\sec^2(2x))$

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Example Question #29 : Chain Rule And Implicit Differentiation

Find the derivative of the function:

$h(t)=(\ln t)^3$

Possible Answers:

$h'(t)=\frac{3(\ln t)^2}{t^2}$

$h'(t)=\frac{-3(\ln t)^2}{t}$

$h'(t)=3(\ln t)^2\cdot t$

$h'(t)=\3(\ln t)^2$

$h'(t)=\frac{3(\ln t)^2}{t}$

Correct answer:

$h'(t)=\frac{3(\ln t)^2}{t}$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

g(t)=t^3

and

$f(t)=\ln t$ .

Since

g'(t)=3t^2

and

$f'(t)=\frac{1}{t}$ ,

we can conclude that

$h'(t)=\frac{d}{dt}g(f(t))=3(f(t))^2\cdot\frac{1}{t}$

and

$h'(t)= \frac{3(\ln t)^2}{t}$

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Example Question #121 : Computation Of The Derivative

Find the derivative of:

$n(x)=5^{7x-3}$

Possible Answers:

$n'(x)=5^{7x-3}\ln(5)$

$n'(x)=5^{7x-3}\ln(7x-3)\cdot 7$

$n'(x)=(7x-3)\ln(5)\cdot 7$

$n'(x)=5^{7x-3}\ln(5)\cdot 7$

$n'(x)=5^{7x-3}\ln(7x-3)$

Correct answer:

$n'(x)=5^{7x-3}\ln(5)\cdot 7$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dx}g(f(x))=g'(f(x))f'(x)$

So in this problem we let

g(x)=5^x

and

f(x)=7x-3 .

Since

$g'(x)=5^x\ln(5)$

and

f'(x)=7 ,

we can conclude that

$n'(x)=\frac{d}{dx}g(f(x))=5^{f(x)}\ln(5)\cdot 7=5^{7x-3}\ln(5)\cdot 7$

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Example Question #421 : Ap Calculus Ab

Find the derivative of:

$p(x)=(7-4x^2)^{200}$

Possible Answers:

$p'(x)=1600x(7-4x^2)^{199}$

$p'(x)=-1600x(7-4x^2)^{199}$

$p'(x)=-800x^2(7-4x^2)^{199}$

$p'(x)=200(7-4x^2)^{199}$

$p'(x)=-1600x(7-4x^2)^{100}$

Correct answer:

$p'(x)=-1600x(7-4x^2)^{199}$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dx}g(f(x))=g'(f(x))f'(x)$

So in this problem we let

$g(x)=x^{200}$

and

f(x)=7-4x^2 .

Since

$g'(x)=200x^{199}$

and

f'(x)=-8x ,

we can conclude that

$p'(x)=\frac{d}{dx}g(f(x))=200(f(x))^{199}\cdot (-8x)$

$p'(x)=-1600x(7-4x^2)^{199}$

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Example Question #31 : Chain Rule And Implicit Differentiation

Find the derivative of:

$h(t)=2\sin\pi t^2$

Possible Answers:

$h'(t)=-4\pi t\cdot\cos\pi t^2$

$h'(t)=4\pi t\cdot\cos\pi t^2$

$h'(t)=2\cos\pi t^2\cdot(t^2+2\pi t)$

$h'(t)=2\pi t\cdot\cos\pi t^2$

$h'(t)=4 \cos\pi t^2$

Correct answer:

$h'(t)=4\pi t\cdot\cos\pi t^2$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

g(t)=2sin(t)

and

$f(t)=\pi t^2$ .

Since

$g'(t)=2\cos(t)$

and

$f'(t)=2\pi t$ ,

we can conclude that

$h'(t)=\frac{d}{dt}g(f(t))=2\cos(f(t))\cdot2\pi t$

and

$h'(t)=4\pi t\cdot\cos\pi t^2$

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Example Question #421 : Ap Calculus Ab

Find the derivative of:

$p(x)=\tan^4 x$

Possible Answers:

$p'(x)=4\tan^3x \cdot \sec^2 x$

Correct answer:

$p'(x)=4\tan^3x \cdot \sec^2 x$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dx}g(f(x))=g'(f(x))f'(x)$

So in this problem we let

g(x)=x^4

and

$f(x)=\tan x$ .

Since

g'(x)=4x^3

and

$f'(x)=\sec^2 x$ ,

we can conclude that

$p'(x)=\frac{d}{dx}g(f(x))=4(f(x))^3\cdot\sec^2 x$

$p'(x)=4\tan^3x \cdot \sec^2 x$

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Example Question #431 : Ap Calculus Ab

Find the derivative of:

$h(t)=\ln(\ln t)$

Possible Answers:

$h'(t)=\frac{1}{t\ln t}$

$h'(t)=\frac{1}{\ln t}+t$

$h'(t)=\frac{t}{\ln t}$

$h'(t)=t\ln t$

$h'(t)=\frac{\ln t}{t}$

Correct answer:

$h'(t)=\frac{1}{t\ln t}$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

$g(t)=\ln(t)$

and

$f(t)=\ln(t)$ .

Since

$g'(t)=\frac{1}{t}$

and

$f'(t)=\frac{1}{t}$ ,

we can conclude that

$h'(t)=\frac{d}{dt}g(f(t))=\frac{1}{f(t)}\cdot\frac{1}{t}=\frac{1}{t\ln t}$

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Example Question #432 : Ap Calculus Ab

Find the derivative of:

$f(x)=\ln(\sin e^x)$

Possible Answers:

$f'(x)=\frac{e^x}{\sin e^x \cos e^x}$

$f'(x)=e^x\tan e^x$

$f'(x)=e^x\cot e^x$

$f'(x)=e^x\cos^2 e^x$

$f'(x)=\frac{\cos e^x}{\sin e^x}$

Correct answer:

$f'(x)=e^x\cot e^x$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dx}g(h(j(x)))=g'(h(j(x)))h'(j(x))j'(x)$

So in this problem we let

$g(x)=\ln x$

and

$h(x)=\sin x$

and

j(x)=e^x .

Since

$g'(x)=\frac{1}{x}$

and

$h'(x)=\cos x$

and

j'(x)=e^x ,

we can conclude that

$f'(x)=\frac{d}{dx}g(h(j(x)))=\frac{1}{h(j(x))}\cdot cos(j(x))\cdot e^x$

$f'(x)=\frac{1}{\sin e^x}\cdot \cos e^x\cdot e^x=e^x\cot e^x$

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Example Question #433 : Ap Calculus Ab

Find dy/dx of:

$y=e^{\cos x}$

Possible Answers:

$\frac{dy}{dx}= =e^{\sin x}\cos x$

$\frac{dy}{dx}= -e^{\cos x}\sin x$

$\frac{dy}{dx}= e^{\cos x}\cos x$

$\frac{dy}{dx}= -e^{\cos x}\ln(\cos x)\cdot\sin x$

$\frac{dy}{dx}= e^{\cos x}\sin x$

Correct answer:

$\frac{dy}{dx}= -e^{\cos x}\sin x$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dx}g(f(x))=g'(f(x))f'(x)$

So in this problem we let

g(x)=e^x

and

$f(x)=\cos x$ .

Since

g'(x)=e^x

and

$f'(x)=-\sin x$ ,

we can conclude that

$\frac{dy}{dx}=\frac{d}{dx}g(f(x))=e^{(f(x))}(-\sin x) = -e^{\cos x}\sin x$

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AP Calculus AB : Computation of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #21 : Chain Rule And Implicit Differentiation

Example Question #28 : Chain Rule And Implicit Differentiation

Example Question #29 : Chain Rule And Implicit Differentiation

Example Question #121 : Computation Of The Derivative

Example Question #421 : Ap Calculus Ab

Example Question #31 : Chain Rule And Implicit Differentiation

Example Question #421 : Ap Calculus Ab

Example Question #431 : Ap Calculus Ab

Example Question #432 : Ap Calculus Ab

Example Question #433 : Ap Calculus Ab

All AP Calculus AB Resources

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