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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #461 : Ap Calculus Ab

Given the function $\small \small \small \small z(x)=\sec ^4 x$ , find its derivative.

Possible Answers:

$\small \small z'(x)=4\sec^4x\tan x$

$\small \small \small z'(x)=4\sec^3x\tan x$

$\small \small \small z'(x)=4\sec^3x$

$\small \small \small z'(x)=4\sec^4x$

Correct answer:

$\small \small z'(x)=4\sec^4x\tan x$

Explanation:

Given the function $\small \small \small \small z(x)=\sec ^4 x$ , we can find its derivative using the chain rule, which states that

$\small [f(g(x))]'=f'(g(x))g'(x)$

where $\small \small \small \small \small \small \small f(x)=x^4$ and $\small \small \small g(x)=\sec x$ for $\small z(x)$ . We have $\small \small \small \small \small \small f'(x)=4x^3$ and $\small \small \small g'(x)=\sec x \tan x$ , which gives us

$\small \small \small \small \small \small z'(x)=4\sec^3x\cdot (\sec x\tan x)=4\sec^4x\tan x$

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Example Question #462 : Ap Calculus Ab

$f(x) = \cos (\sqrt{x^{2}- 2x} )$

Find f'(x) .

Possible Answers:

$f'(x) =- \frac{ ( x-1) \sin \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) = \frac{ ( x-1) \cos \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) = \frac{ ( x-1) \sin \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) = \frac{ ( x-1) \csc \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

$f'(x) =- \frac{ ( x-1) \cos \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

Correct answer:

$f'(x) =- \frac{ ( x-1) \sin \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

Explanation:

Let $u(x) = x^{2}- 2x$

Then

$f(x) = \cos (\sqrt{x^{2}- 2x} )$

can be rewritten as

$f(x) = \cos (\sqrt{u } )$

Let $v (u) = \sqrt{u} =u^{\frac{1}{2}}$

The function can now be rewritten as

$f(x) = \cos v$

Applying the chain rule twice:

$\frac{\mathrm{d} f}{\mathrm{d} x} = \frac{\mathrm{d} f}{\mathrm{d} v} \cdot \frac{\mathrm{d} v}{\mathrm{d} u} \cdot \frac{\mathrm{d} u}{\mathrm{d} x}$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin v \cdot \frac{1}{2} u ^{-\frac{1}{2}} \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin u^{\frac{1}{2}} \cdot \frac{1}{2} u ^{-\frac{1}{2}} \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin \sqrt{u} \cdot \frac{1}{2\sqrt{u} } \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} =-\sin \sqrt{x^{2}- 2x} \cdot \frac{1}{2\sqrt{x^{2}- 2x} } \cdot (2x-2)$

$\frac{\mathrm{d} f}{\mathrm{d} x} = \frac{-\sin \sqrt{x^{2}- 2x} \cdot (2x-2)}{2\sqrt{x^{2}- 2x} }$

$\frac{\mathrm{d} f}{\mathrm{d} x} =- \frac{ ( x-1) \sin \sqrt{x^{2}- 2x}}{\sqrt{x^{2}- 2x} }$

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Example Question #463 : Ap Calculus Ab

Find the derivative of the function

$f(x)=2\sin(x^2)$

Possible Answers:

$f'(x)=4x\sin(x^2)$

$f'(x)=-4x\cos(x^2)$

$f'(x)=4x\cos(2x)$

$f'(x)=4x\cos(x^2)$

Correct answer:

$f'(x)=4x\cos(x^2)$

Explanation:

To find the derivative of the function, you must apply the chain rule, which is as follows:

$\frac{\mathrm{d} }{\mathrm{d} x}(p(g(x)))=p'(g(x))(g'(x))$

Using the function from the problem statement, we have that

g(x)=x^2 and $p(g(x))=2\sin(x^2)$

Following the rule, we get

$f'(x)=4x\cos(x^2)$

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Example Question #464 : Ap Calculus Ab

Find the derivative of the function

$w(x)=\sin(\tan{x})$

Possible Answers:

$w'(x)=\sec^2(x)\cos(\tan{x})$

$w'(x)=\sec^2(x)\sin(\tan{x})$

$w'(x)=\sec^2(x)-\cos(\tan{x})$

$w'(x)=\sec(x)\cos(\tan{x})$

Correct answer:

$w'(x)=\sec^2(x)\cos(\tan{x})$

Explanation:

To find the derivative of the function, you must apply the chain rule, which is as follows:

$\frac{\mathrm{d} }{\mathrm{d} x}(p(g(x)))=p'(g(x))(g'(x))$

Using the function from the problem statement, we have that

$g(x)=\tan(x)$ and $p(g(x))=\sin(\tan(x))$

Following the rule, we get

$w'(x)=\sec^2(x)\cos(\tan(x))$

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Example Question #465 : Ap Calculus Ab

$f(x) = \cos (x+ \ln x)$

Find f'(x) .

Possible Answers:

None of the other choices gives the correct response.

$f'(x)= \frac{(-x- 1)\sin \left ( x - \ln x \right )}{x}$

$f'(x)= \frac{(-x- 1)\sin \left ( x + \ln x \right )}{x}$

$f'(x)= \frac{(x+1)\sin \left ( x + \ln x \right )}{x}$

$f'(x)= \frac{(x+1)\sin \left ( x - \ln x \right )}{x}$

Correct answer:

$f'(x)= \frac{(-x- 1)\sin \left ( x + \ln x \right )}{x}$

Explanation:

Let $u(x) = x + \ln x$

Then

$f(x) = \cos u$

and

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos (u(x))$

Apply the chain rule:

$f'(x)= \frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}( x + \ln x) \cdot( - \sin u)$

Substitute back for :

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}( x + \ln x) \cdot( - \sin \left ( x + \ln x \right ))$

Apply the sum rule:

$f'(x)=\left ( \frac{\mathrm{d} }{\mathrm{d} x}( x ) + \frac{\mathrm{d} }{\mathrm{d} x}( \ln x) \right )\cdot( - \sin \left ( x + \ln x \right ))$

$f'(x)=\left ( 1 + \frac{1}{x} \right )\cdot( - \sin \left ( x + \ln x \right ))$

After some simple algebra:

$f'(x)=\left ( \frac{x+ 1}{x} \right )\cdot( - \sin \left ( x + \ln x \right ))$

$f'(x)= \frac{(-x- 1)\sin \left ( x + \ln x \right )}{x}$

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Example Question #466 : Ap Calculus Ab

is a function of . Solve for in this differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{xy}{x^{2}+ 2}$

Possible Answers:

$y =C e^{\sqrt{ \frac{1}{3}x^{3}+ 2x}}$

$y =\frac{C\sqrt{ x^{2}+ 2} }{ x^{2}+ 2}$

$y = \frac{1}{3}x^{3}+ 2x+ C$

$y =C \sqrt{ x^{2}+ 2}$

$y =C e^{\sqrt{ x^{2}+ 2}}$

Correct answer:

$y =C \sqrt{ x^{2}+ 2}$

Explanation:

The expressions with can be separated from those with by multiplying both sides by $\frac{1}{y}\cdot \mathrm{d} x$ :

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{1}{y}\cdot \mathrm{d} x = \frac{xy}{x^{2}+ 2} \cdot \frac{1}{y}\cdot \mathrm{d} x$

$\frac{\mathrm{d} y}{y} = \frac{x \mathrm{\; d} x}{x^{2}+ 2}$

Find the indefinite integral of both sides:

$\int \frac{ \mathrm{d} y }{y}=\int \frac{x \mathrm{ \; d} x}{x^{2}+ 2 }$

$\ln y =\int \frac{x \mathrm{\; d} x}{x^{2}+ 2 }$

Set $u = x^{2}+ 2$ . Then $\mathrm{d} u = 2x \; \mathrm{ d} x$ , or $x \; \mathrm{ d} x =\frac{1}{2} \mathrm{d} u$ , and

$\ln y =\frac{1}{2}\int \frac{ { d} u}{u }$

$\ln y =\frac{1}{2} ( \ln u + C)$

$\ln y =\frac{1}{2} \ln u + C_{1}$

$\ln y = \ln \sqrt{u} + C_{1}$

Substitute back:

$\ln y = \ln \sqrt{ x^{2}+ 2} + C_{1}$ ;

Raise to both powers:

$\ln y = \ln \sqrt{ x^{2}+ 2} + C_{1}$

$e^{ \ln y }=e^{ \ln \sqrt{ x^{2}+ 2} + C_{1}}$

$y =C_{2}e^{ \ln \sqrt{ x^{2}+ 2} }$

$y =C_{2} \sqrt{ x^{2}+ 2}$ .

The correct choice is $y =C \sqrt{ x^{2}+ 2}$

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Example Question #467 : Ap Calculus Ab

Find the derivative using the chain rule.

$\cos (3x^2)$

Possible Answers:

$-6x\cos (3x^2)$

$-6x\sin (3x^2)$

$6x\cos(3x^2)$

$6x\sin (3x^2)$

Correct answer:

$-6x\sin (3x^2)$

Explanation:

Use the chain rule to find the derivative: $\frac{d}{dx}f(g(x))=g'(x)f'(g(x))$

Thus, $\frac{d}{dx}\cos (3x^2)=-6x\sin (3x^2)$

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Example Question #71 : Chain Rule And Implicit Differentiation

Find the derivative using the chain rule.

$\tan (5x)$

Possible Answers:

$5\sec^2(5x)$

$5\sec (5x)$

$\sec^2(5)$

$\sec^2(5x)$

Correct answer:

$5\sec^2(5x)$

Explanation:

Use the chain rule to find the derivative.

$\frac{d}{dx}\tan (5x)=5\sec^2(5x)$

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Example Question #72 : Chain Rule And Implicit Differentiation

Find the derivative using the chain rule.

$\sin (-3x)$

Possible Answers:

$3\sin (3x)$

$-3\sin (-3x)$

$-3\cos (-3x)$

$3\cos (3x)$

Correct answer:

$-3\cos (-3x)$

Explanation:

Use the chain rule to find the derivative.

$\frac{d}{dx}\sin (-3x)=-3\cos (-3x)$

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Example Question #73 : Chain Rule And Implicit Differentiation

Find the derivative using the chain rule.

$\cos (-2x)$

Possible Answers:

$-2\sin (-2x)$

$2\cos (-2x)$

$2\sin (-2x)$

$-2\cos (-2x)$

Correct answer:

$2\sin (-2x)$

Explanation:

Use the chain rule to find the derivative.

$\frac{d}{dx}\cos (-2x)=2\sin(-2x)$

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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 #461 : Ap Calculus Ab

Example Question #462 : Ap Calculus Ab

Example Question #463 : Ap Calculus Ab

Example Question #464 : Ap Calculus Ab

Example Question #465 : Ap Calculus Ab

Example Question #466 : Ap Calculus Ab

Example Question #467 : Ap Calculus Ab

Example Question #71 : Chain Rule And Implicit Differentiation

Example Question #72 : Chain Rule And Implicit Differentiation

Example Question #73 : Chain Rule And Implicit Differentiation

All AP Calculus AB Resources

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