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AP Calculus AB : AP Calculus AB

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

Example Questions

Example Question #201 : Derivatives

Find the derivative: $\frac{dy}{dx}$

$x\sin(y)+y^2x=4x$

Possible Answers:

$\frac{dy}{dx}=\frac{1-xy^2+\cos(y)}{y\sin(y)+1}$

$\frac{dy}{dx}=\frac{y-\cos(y)}{x\sin(y)+y}$

$\frac{dy}{dx}=\frac{2+yx^2+\sin(y)}{y\cos(y)+1}$

$\frac{dy}{dx}=\frac{1-xy^2+x\cos(y)}{x\sin(y)+y}$

$\frac{dy}{dx}=\frac{4-y^2-\sin(y)}{x\cos(y)+2xy}$

Correct answer:

$\frac{dy}{dx}=\frac{4-y^2-\sin(y)}{x\cos(y)+2xy}$

Explanation:

$x\sin(y)+y^2x=4x$

Implicit differentiation is useful when we have an equation containing a dependent variable , and and an independent variable , but we cannot solve to write explicitly in terms of .

Differentiate both sides simultaneously:

$\underset{Product\: \: Rule \: \: 1st\: \: term}{ \underbrace{\sin(y)\left(\frac{d}{dx}x \right )+x\left(\frac{d}{dx}\sin (y) \right ) }} + \underset{Product\: \: Rule \: \: 2nd\: \: term}{ \underbrace{y^2\left(\frac{d}{dx}x\right)+x\left(\frac{d}{dx} y^2\right ) }}=\frac{d}{dx}4x$

The chain rule must be applied to the following derivatives:

$\frac{d}{dx}\sin (y)\: \; \; \; \; \; \; (1)\: \: \: \: \: \: \: \:\frac{d}{dx}y^2\: \: \: \: \: \: \:(2)$

Noting the is implied to be a function of , we have to first differentiate with respect to and then differentiate with respect to Since we do not have an explicit definition for , we simply express its' derivative in the equation as $\frac{dy}{dx}$ , treating it as an "unknown," which needs to be solved for algebraically at a later step.

For (1):

$\frac{d}{dx}\sin (y)=\left( \frac{d}{dy}\sin (y) \right )\frac{dy}{dx}=\left(\cos(y)\right )\frac{dy}{dx}$

For (2):

$\frac{d}{dx}y^2=\left(\frac{d}{dy}y^2\right)\frac{dy}{dx}=2y\frac{dy}{dx}$

Using these derivatives the main equation becomes:

$\sin (y)+x\cos (y)\frac{dy}{dx} +y^2 + 2xy\frac{dy}{dx}=4$

Now we simply solve for $\frac{dy}{dx}$ and simplify as much as possible. Isolate all terms with a derivative onto one side of the equation and factor:

$x\cos(y)\frac{dy}{dx}+2xy\frac{dy}{dx}=4-y^2-\sin(y)$

$\frac{dy}{dx}[x\cos(y)+2xy]=4-y^2-\sin(y)$

$\frac{dy}{dx}=\frac{4-y^2-\sin(y)}{x\cos(y)+2xy}$

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Example Question #23 : Chain Rule And Implicit Differentiation

Find the derivative of the function:

$h(x)=e^{2x^3}$

Possible Answers:

$h'(x)=e^{2x^3}\cdot 2x^3$

$h'(x)=e^{2x^3}\cdot ln(e) \cdot 2x^3$

$h'(x)=e^{2x^3}$

$h'(x)=e^{6x^2}\cdot 6x^2$

$h'(x)=e^{2x^3}\cdot 6x^2$

Correct answer:

$h'(x)=e^{2x^3}\cdot 6x^2$

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)=2x^3 .

Since

g'(x)=e^x

and

f'(x)=6x^2 ,

we can conclude that

$h'(x)=\frac{d}{dx}g(f(x))=e^{f(x)}\cdot 6x^2=e^{2x^3}\cdot 6x^2$

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Example Question #22 : Chain Rule And Implicit Differentiation

Find the derivative of:

$h(x)=(5-5x)^{3/2}$

Possible Answers:

$h'(x)=\frac{-15}{2}(5-5x)^{-1/2}$

$h'(x)=\frac{3}{2}(5-5x)^{1/2}$

$h'(x)=\frac{-15}{2}(5-5x)^{3/2}$

$h'(x)=\frac{3}{2}(5-5x)^{3/2}$

$h'(x)=\frac{-15}{2}(5-5x)^{1/2}$

Correct answer:

$h'(x)=\frac{-15}{2}(5-5x)^{1/2}$

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

$g(x)=x^{3/2}$

and

f(x)=5-5x .

Since

$g'(x)=\frac{3}{2}x^{1/2}$

and

f'(x)=-5 ,

we can conclude that

$h'(x)=\frac{d}{dx}g(f(x))=\frac{3}{2}(f(x))^{1/2}\cdot -5$

$=\frac{3}{2}(5-5x)^{1/2}\cdot -5=\frac{-15}{2}(5-5x)^{1/2}$

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Example Question #121 : Computation Of The Derivative

Find the derivative of:

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

Possible Answers:

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

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

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

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

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

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 #23 : Chain Rule And Implicit Differentiation

Find the derivative of:

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

Possible Answers:

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

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

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

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

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

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 #24 : Chain Rule And Implicit Differentiation

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

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

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 #24 : Chain Rule And Implicit Differentiation

Find the derivative of:

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

Possible Answers:

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

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

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

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

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

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 #31 : Chain Rule And Implicit Differentiation

Find the derivative of:

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

Possible Answers:

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

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

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

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

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

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 #32 : Chain Rule And Implicit Differentiation

Find the derivative of:

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

Possible Answers:

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

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

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

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

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

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 #33 : Chain Rule And Implicit Differentiation

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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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #201 : Derivatives

Example Question #23 : Chain Rule And Implicit Differentiation

Example Question #22 : Chain Rule And Implicit Differentiation

Example Question #121 : Computation Of The Derivative

Example Question #23 : Chain Rule And Implicit Differentiation

Example Question #24 : Chain Rule And Implicit Differentiation

Example Question #24 : Chain Rule And Implicit Differentiation

Example Question #31 : Chain Rule And Implicit Differentiation

Example Question #32 : Chain Rule And Implicit Differentiation

Example Question #33 : Chain Rule And Implicit Differentiation

All AP Calculus AB Resources

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