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Calculus 3 : Calculus 3

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #91 : Derivatives

Find the derivative $\small f'(x)$ of the function $\small \small \small \small \small \small \small \small f(x)=\frac{x^5}{5!}$

Possible Answers:

$\small \small \small \small \small f'(x)=\frac{x^6}{6!}$

$\small \small \small \small f'(x)=\frac{x^4}{4!}$

$\small \small \small \small \small f'(x)=\frac{x^5}{4!}$

$\small \small \small \small \small f'(x)=\frac{x^3}{3!}$

Correct answer:

$\small \small \small \small f'(x)=\frac{x^4}{4!}$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small \small \small \small \small \small \small \small f(x)=\frac{x^5}{5!}$ by using the power rule for derivatives:

$\small \frac{d}{dx}x^{a}=ax^{a-1}$

with $\small \small \small \small \small \small a=5$ to get

$\small \small \small f'(x)=5\cdot \frac{x^4}{5!}=\frac{5}{5\cdot 4!}x^4=\frac{x^4}{4!}$

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Example Question #93 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small \small f(x)=\frac{x^7}{7!}$

Possible Answers:

$\small \small \small \small \small \small f'(x)=\frac{x^6}{8!}$

$\small \small \small \small \small \small f'(x)=\frac{x^6}{7!}$

$\small \small \small \small \small \small f'(x)=\frac{x^7}{7!}$

$\small \small \small \small \small f'(x)=\frac{x^6}{6!}$

Correct answer:

$\small \small \small \small \small f'(x)=\frac{x^6}{6!}$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small \small f(x)=\frac{x^7}{7!}$ by using the power rule for derivatives:

$\small \frac{d}{dx}x^{a}=ax^{a-1}$

with $\small \small \small \small \small \small \small a=7$ to get

$\small \small \small \small f'(x)=7\cdot \frac{x^6}{7!}=\frac{7}{7\cdot 6!}x^6=\frac{x^6}{6!}$

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Example Question #94 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small \small \small f(x)=(x^2+x+1)^5$

Possible Answers:

$\small \small \small f'(x)=5(x^2+x+1)^4$

$\small \small \small f'(x)=5(2x+1)^4$

$\small \small f'(x)=5(2x+1)(x^2+x+1)^4$

$\small \small \small f'(x)=5(2x+1)(x^2+x+1)^5$

Correct answer:

$\small \small f'(x)=5(2x+1)(x^2+x+1)^4$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small \small \small f(x)=(x^2+x+1)^5$ by using the power rule for derivatives:

$\small \frac{d}{dx}x^{a}=ax^{a-1}$

with $\small \small \small \small \small \small \small \small a=5$ with the chain rule to get

$\small \small \small \small \small f'(x)=(2x+1)\cdot 5(x^2+x+1)^4=5(2x+1)(x^2+x+1)^4$

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Example Question #95 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small \small \small \small f(x)=(x^4+x^{2.4}+1)^3$

Possible Answers:

$\small \small \small \small \small f'(x)=3(4x^3+2.4x^{1.4})(x^4+x^{2.4}+1)^3$

$\small \small \small \small \small \small f'(x)=2(4x^3+2.4x)(x^4+x^{2.4}+1)^3$

$\small \small \small \small f'(x)=3(4x^3+2.4x^{1.4})(x^4+x^{2.4}+1)^2$

Correct answer:

$\small \small \small \small f'(x)=3(4x^3+2.4x^{1.4})(x^4+x^{2.4}+1)^2$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small \small \small \small f(x)=(x^4+x^{2.4}+1)^3$ by using the power rule for derivatives:

$\small \frac{d}{dx}x^{a}=ax^{a-1}$

with $\small \small \small \small \small \small \small \small \small a=3$ with the chain rule to get

$\small \small \small f'(x)=(4x^3+2.4x^{1.4})\cdot 3(x^4+x^{2.4}+1)^2=3(4x^3+2.4x^{1.4})(x^4+x^{2.4}+1)^2$

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Example Question #96 : Calculus Review

Find the derivative of the function $h(x)=x^2e^{5x}$

Possible Answers:

$h'(x)=(5+2x)e^{2x}$

$h'(x)=(2x^2+5x)e^{2x}$

$h'(x)=(2x+5x^2)e^{2x}$

$h'(x)=(2+5x)e^{2x}$

Correct answer:

$h'(x)=(2x+5x^2)e^{2x}$

Explanation:

To find the derivative of the function $h(x)=x^2e^{5x}$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with f(x)=x^2 and $g(x)=e^{5x}$

so we get

$h'(x)=2x\cdot e^{5x}+x^2\cdot 5e^{5x}=(2x+5x^2)e^{2x}$

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Example Question #282 : Calculus 3

Find the derivative of the function $h(x)=x^5\cos 3x$

Possible Answers:

$h'(x)=5x^4\cos 3x-3x^5\sin 3x$

$h'(x)=-5x^4\cos 3x+3x^5\sin 3x$

$h'(x)=5x^4\cos 3x+3x^5\sin 3x$

$h'(x)=5x^4\sin 3x+3x^5\cos 3x$

Correct answer:

$h'(x)=5x^4\cos 3x-3x^5\sin 3x$

Explanation:

To find the derivative of the function $h(x)=x^5\cos 3x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with f(x)=x^5 and $g(x)=\cos 3x$

so we get

$h'(x)=5x^4\cdot \cos 3x+x^5\cdot (-3\sin 3x)=5x^4\cos 3x-3x^5\sin 3x$

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Example Question #98 : Calculus Review

Find the derivative of the function $h(x)=x^{2.5}(1+2x)^3$

Possible Answers:

$h'(x)=(2.5\sqrt{x}(1+2x)+6x^{2.5})(1+2x)^2$

$h'(x)=(2.5\sqrt{x}(1+2x)+3x^{2.5})(1+2x)^2$

$h'(x)=(6\sqrt{x}(1+2x)+2.5x^{2.5})(1+2x)^2$

$h'(x)=(-2.5\sqrt{x}(1+2x)+6x^{2.5})(1+2x)^2$

Correct answer:

$h'(x)=(2.5\sqrt{x}(1+2x)+6x^{2.5})(1+2x)^2$

Explanation:

To find the derivative of the function $h(x)=x^{2.5}(1+2x)^3$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with $f(x)=x^{2.5}$ and g(x)=(1+2x)^3

so we get

$h'(x)=2.5\sqrt{x}\cdot (1+2x)^3+x^{2.5}\cdot 2\cdot 3(1+2x)^2$

$=(2.5\sqrt{x}(1+2x)+6x^{2.5})(1+2x)^2$

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Example Question #99 : Calculus Review

Find the derivative of the function $h(x)=xe^{2x}$

Possible Answers:

$h'(x)=(1+x)e^{2x}$

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

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

$h'(x)=(1+2x)e^{2x}$

Correct answer:

$h'(x)=(1+2x)e^{2x}$

Explanation:

To find the derivative of the function $h(x)=xe^{2x}$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with f(x)=x and $g(x)=e^{2x}$

so we get

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

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Example Question #100 : Calculus Review

Find the derivative of the function $h(x)=\sin x\cos x$

Possible Answers:

$h'(x)=\cos^2x-\sin^2x$

$h'(x)=\cos^2x\sin^2x$

h'(x)=1

$h'(x)=\cos x\sin x$

Correct answer:

$h'(x)=\cos^2x-\sin^2x$

Explanation:

To find the derivative of the function $h(x)=\sin x\cos x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with $f(x)=\sin x$ and $g(x)=\cos x$

so we get

$h'(x)=\cos x\cdot \cos x+\sin x\cdot (-\sin x)=\cos^2x-\sin^2x$

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Example Question #101 : Derivatives

Find the derivative of the function $h(x)=\sec x\tan x$

Possible Answers:

$h'(x)=\sec^3 x\tan x$

$h'(x)=\sec x\tan^3x+\sec^3x$

$h'(x)=\sec x\tan^2x+\sec^3x$

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

Correct answer:

$h'(x)=\sec x\tan^2x+\sec^3x$

Explanation:

To find the derivative of the function $h(x)=\sec x\tan x$ , we use the product rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

with $f(x)=\sec x$ and $g(x)=\tan x$

so we get

$h'(x)=\sec x \tan x\cdot \tan x+\sec x\cdot \sec^2 x=\sec x\tan^2x+\sec^3x$

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #91 : Derivatives

Example Question #93 : Calculus Review

Example Question #94 : Calculus Review

Example Question #95 : Calculus Review

Example Question #96 : Calculus Review

Example Question #282 : Calculus 3

Example Question #98 : Calculus Review

Example Question #99 : Calculus Review

Example Question #100 : Calculus Review

Example Question #101 : Derivatives

All Calculus 3 Resources

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