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AP Calculus AB : Computation of the Derivative

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Example Questions

Example Question #237 : Derivatives

Given the function $\small \small \small \small z(x)=e^{x^{-1}}$ , find its derivative.

Possible Answers:

$\small \small \small \small \small z'(x)=-\frac{e^{x^{-1}}}{x}$

$\small \small \small \small \small z'(x)=e^{-x^{-2}}$

$\small \small \small \small z'(x)=-\frac{e^{x^{-1}}}{x^2}$

$\small \small \small \small \small z'(x)=\frac{e^{x^{-1}}}{x}$

Correct answer:

$\small \small \small \small z'(x)=-\frac{e^{x^{-1}}}{x^2}$

Explanation:

Given the function $\small \small \small \small z(x)=e^{x^{-1}}$ , we can find its derivative using the chain rule, which states that

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

where $\small \small \small f(x)=e^{x}$ and $\small \small \small g(x)=x^{-1}$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small g'(x)=-x^{-2}$ , which gives us

$\small \small \small z'(x)=e^{x^{-1}}\cdot (-x^{-2})=-\frac{e^{x^{-1}}}{x^2}$

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Example Question #238 : Derivatives

Given the function $\small \small \small z(x)=e^{-x+x^3}$ , find its derivative.

Possible Answers:

$\small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-x+x^{3})$

$\small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+x^{3})$

$\small \small \small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+3x^{2})$

$\small \small \small \small \small z'(x)=e^{-1+3x^2}\cdot (-1+x^{3})$

Correct answer:

$\small \small \small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+3x^{2})$

Explanation:

Given the function $\small \small \small z(x)=e^{-x+x^3}$ , we can find its derivative using the chain rule, which states that

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

where $\small \small \small f(x)=e^{x}$ and $\small \small \small \small g(x)=-x+x^3$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small \small g'(x)=-1+3x^2$ , which gives us

$\small \small \small \small \small z'(x)=e^{-x+x^3}\cdot (-1+3x^{2})$

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Example Question #239 : Derivatives

Given the function $\small \small \small \small \small z(x)=e^{\cos x+\sin x}$ , find its derivative.

Possible Answers:

$\small \small z'(x)=e^{\cos x+\sin x}\cdot (-\sin x+\cos x)$

$\small \small \small z'(x)=e^{\cos x-\sin x}\cdot (-\sin x+\cos x)$

$\small \small \small z'(x)=e^{\cos x+\sin x}$

$\small \small \small z'(x)=e^{\cos x+\sin x}\cdot (\sin x+\cos x)$

Correct answer:

$\small \small z'(x)=e^{\cos x+\sin x}\cdot (-\sin x+\cos x)$

Explanation:

Given the function $\small \small \small \small \small z(x)=e^{\cos x+\sin x}$ , we can find its derivative using the chain rule, which states that

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

where $\small \small \small \small f(x)=e^{x}$ and $\small \small \small \small \small \small g(x)=\cos x+\sin x$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small \small \small \small \small g'(x)=-\sin x+\cos x$ , which gives us

$\small \small z'(x)=e^{\cos x+\sin x}\cdot (-\sin x+\cos x)$

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Example Question #240 : Derivatives

Given the function $\small \small \small \small z(x)=e^{e^x-x^5}$ , find its derivative.

Possible Answers:

$\small \small \small \small \small \small \small z'(x)=e^{e^x-5x^4}$

$\small \small \small \small \small \small \small \small z'(x)=e^{e^x-5x^4}\cdot (e^x-x^5)$

$\small \small \small \small \small \small \small z'(x)=e^{e^x-x^5}\cdot (e^x-5x^4)$

$\small \small \small \small \small \small \small z'(x)=e^{e^x-5x^4}\cdot (e^x-5x^4)$

Correct answer:

$\small \small \small \small \small \small \small z'(x)=e^{e^x-x^5}\cdot (e^x-5x^4)$

Explanation:

Given the function $\small \small \small \small z(x)=e^{e^x-x^5}$ , we can find its derivative using the chain rule, which states that

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

where $\small \small \small \small f(x)=e^{x}$ and $\small \small \small \small \small g(x)=e^x-x^5$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small \small \small g'(x)=e^x-5x^4$ , which gives us

$\small \small \small \small \small \small \small z'(x)=e^{e^x-x^5}\cdot (e^x-5x^4)$

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Example Question #241 : Derivatives

Given the function $\small \small \small \small \small \small z(x)=e^{1+x+x^2}$ , find its derivative.

Possible Answers:

$\small \small \small \small z'(x)=e^{1+2x}$

$\small \small \small \small z'(x)=e^{1+2x}\cdot(1+2x)$

$\small \small \small \small z'(x)=e^{1+x+x^2}$

$\small \small \small z'(x)=e^{1+x+x^2}\cdot(1+2x)$

Correct answer:

$\small \small \small z'(x)=e^{1+x+x^2}\cdot(1+2x)$

Explanation:

Given the function $\small \small \small \small \small \small z(x)=e^{1+x+x^2}$ , we can find its derivative using the chain rule, which states that

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

where $\small \small \small \small f(x)=e^{x}$ and $\small \small \small \small \small \small \small g(x)=1+x+x^2$ for $\small z(x)$ . We have $\small \small f'(x)=e^x$ and $\small \small \small \small \small \small \small g'(x)=1+2x$ , which gives us

$\small \small \small z'(x)=e^{1+x+x^2}\cdot(1+2x)$

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Example Question #242 : Derivatives

Given the function $\small \small z(x)=\cos (5-x^{10})$ , find its derivative.

Possible Answers:

$\small \small \small \small \small \small z'(x)=10x^{9}\cos(5-x^{10})$

$\small \small \small \small \small \small \small z'(x)=-10x^{9}\cos(5-x^{10})$

$\small \small \small \small \small \small z'(x)=-10x^{9}\sin(5-x^{10})$

$\small \small \small \small \small z'(x)=10x^{9}\sin(5-x^{10})$

Correct answer:

$\small \small \small \small \small z'(x)=10x^{9}\sin(5-x^{10})$

Explanation:

Given the function $\small \small z(x)=\cos (5-x^{10})$ , we can find its derivative using the chain rule, which states that

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

where $\small \small \small \small \small f(x)=\cos x$ and $\small \small \small \small \small \small \small \small \small g(x)=5-x^{10}$ for $\small z(x)$ . We have $\small \small \small \small f'(x)=-\sin x$ and $\small \small \small \small \small \small \small \small \small \small g'(x)=-10x^{9}$ , which gives us

$\small \small \small \small z'(x)=[-\sin (5-x^{10})][-10x^9]=10x^{9}\sin(5-x^{10})$

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Example Question #243 : Derivatives

Given the function $\small \small \small z(x)=\cos ^3 x$ , find its derivative.

Possible Answers:

$\small \small \small \small \small \small z'(x)=-3\cos^2x\cdot \sin x$

$\small \small \small \small \small \small \small z'(x)=3\cos^2x\cdot \sin x$

$\small \small \small \small \small \small \small z'(x)=3\sin^2x\cdot \cos x$

$\small \small \small \small \small \small \small z'(x)=-3\cos^2x$

Correct answer:

$\small \small \small \small \small \small z'(x)=-3\cos^2x\cdot \sin x$

Explanation:

Given the function $\small \small \small z(x)=\cos ^3 x$ , we can find its derivative using the chain rule, which states that

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

where $\small \small \small \small \small \small f(x)=x^3$ and $\small \small g(x)=\cos x$ for $\small z(x)$ . We have $\small \small \small \small \small f'(x)=3x^2$ and $\small \small g'(x)=-\sin x$ , which gives us

$\small \small \small \small \small z'(x)=3\cos^2x\cdot (-\sin x)$

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Example Question #244 : Derivatives

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

Possible Answers:

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

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

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

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

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 #245 : Derivatives

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

Find f'(x) .

Possible Answers:

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

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

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

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

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

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 #246 : Derivatives

Find the derivative of the function

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

Possible Answers:

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

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

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

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

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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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 #237 : Derivatives

Example Question #238 : Derivatives

Example Question #239 : Derivatives

Example Question #240 : Derivatives

Example Question #241 : Derivatives

Example Question #242 : Derivatives

Example Question #243 : Derivatives

Example Question #244 : Derivatives

Example Question #245 : Derivatives

Example Question #246 : Derivatives

All AP Calculus AB Resources

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