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Calculus 3 : Calculus Review

Study concepts, example questions & explanations for Calculus 3

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

Example Questions

Example Question #71 : Derivatives

What is the derivative of $\large x^2+\frac {1}{x}$ ?

Possible Answers:

$\large 2x+\frac {1}{x^2}$

$\large 2x-\frac {1}{x^2}$

$\large \frac {x^3}{3}-\ln(x)$

None of the Above

$\large \frac {x^3}{3}+\ln(x)$

Correct answer:

$\large 2x-\frac {1}{x^2}$

Explanation:

Take the derivative of the first term:

$\large \large (x^2)'=2x$
Rewrite the other term with a(n) negative exponent:

$\large \bigg(\large \frac {1}{x}\bigg)'=x^{-1}$

Now, take the derivative of the re-written term:

$\large (x^{-1})'=-x^{-2}\implies \frac {-1}{x^2}$
Put both parts together:

$\large \large 2x-\frac {1}{x^2}$

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Example Question #71 : Derivatives

Find the derivative $\small f'(x)$ given the function $\small f(x)=2^x$

Possible Answers:

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

$\small \small \small f'(x)=2^x$

$\small f'(x)=2^x\cdot \ln 2$

$\small \small f'(x)=2^x\cdot e^2$

Correct answer:

$\small f'(x)=2^x\cdot \ln 2$

Explanation:

We can find the derivative $\small f'(x)$ given the function $\small f(x)=2^x$ by rewriting $\small f$ as

$\small \small f(x)=2^x=e^{\ln 2^x}$

and using the properties of logarithms ( $\small \ln a^b=b\ln a$ in particular) to get

$\small \small \small f(x)=2^x=e^{\ln 2^x}=e^{x\ln 2}$

so now we can use the chain rule

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

with $\small g(x)=e^x$ and $\small h(x)=x\cdot \ln 2$ to get

$\small f'(x)=e^{x\ln 2}\cdot \ln 2=2^x\cdot \ln 2$

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Example Question #72 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small f(x)=\pi x^{\sqrt{2}}$ .

Possible Answers:

$\small f'(x)=\sqrt{2}\pi x^{\sqrt{2}-1}$

$\small f'(x)=(\sqrt{2}-1)\pi x^{\sqrt{2}-1}$

$\small f'(x)=\sqrt{2}\pi x^{\sqrt{2}}$

$\small f'(x)=\sqrt{2}\pi x^{\sqrt{2}+1}$

Correct answer:

$\small f'(x)=\sqrt{2}\pi x^{\sqrt{2}-1}$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small f(x)=\pi x^{\sqrt{2}}$ by using the power rule for derivatives:

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

with $a=\sqrt{2}$ to get

$\small f'(x)=\pi\cdot \sqrt{2}x^{\sqrt{2}-1}=\sqrt{2}\pi x^{\sqrt{2}-1}$

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Example Question #73 : Calculus Review

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

Possible Answers:

$\small f'(x)=(\sqrt{3}+3)x^{\sqrt{3}+3}$

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

$\small f'(x)=(\sqrt{3}+3)x^{\sqrt{3}+2}$

$\small f'(x)=(\sqrt{3}+2)x^{\sqrt{3}+2}$

Correct answer:

$\small f'(x)=(\sqrt{3}+3)x^{\sqrt{3}+2}$

Explanation:

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

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

with $a=\sqrt{3}+3$ to get

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

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Example Question #74 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small f(x)=x^{\frac{20}{3}}$

Possible Answers:

$\small f'(x)=\frac{17}{3}x^{\frac{17}{3}}$

$\small f'(x)=\frac{20}{3}x^{\frac{17}{3}}$

$\small f'(x)=\frac{20}{3}x^{\frac{20}{3}}$

$\small f'(x)=\frac{17}{3}x^{\frac{20}{3}}$

Correct answer:

$\small f'(x)=\frac{20}{3}x^{\frac{17}{3}}$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small f(x)=x^{\frac{20}{3}}$ by using the power rule for derivatives:

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

with $a=\frac{20}{3}$ to get

$\small f'(x)=\frac{20}{3}x^{\frac{20}{3}-1}=\frac{20}{3}x^{\frac{17}{3}}$

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Example Question #71 : Derivatives

Find the derivative $\small f'(x)$ of the function $\small f(x)=\frac{4}{5}x^{\frac{5}{4}}$

Possible Answers:

$\small f'(x)=x^{\frac{1}{4}}$

$\small f'(x)=\frac{1}{4}x^{\frac{1}{4}}$

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

$\small f'(x)=\frac{5}{4}x^{\frac{1}{4}}$

Correct answer:

$\small f'(x)=x^{\frac{1}{4}}$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small f(x)=\frac{4}{5}x^{\frac{5}{4}}$ by using the power rule for derivatives:

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

with $a=\frac{5}{4}$ to get

$\small f'(x)=\frac{4}{5}\cdot\frac{5}{4}x^{\frac{5}{4}-1}=x^{\frac{1}{4}}$

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Example Question #76 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small f(x)=(1-\pi)x^{\frac{e}{2}}$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small f(x)=(1-\pi)x^{\frac{e}{2}}$ by using the power rule for derivatives:

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

with $a=\frac{e}{2}$ to get

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

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Example Question #77 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small f(x)=(1-e)x^{\frac{\pi}{3}}$

Possible Answers:

$\small f'(x)=\frac{\pi(1-e)}{3}x^{\frac{\pi}{3}}$

$\small f'(x)=\frac{\pi(1-e)}{3}x^{\frac{\pi}{3}-1}$

$\small f'(x)=\frac{3(1-e)}{\pi}x^{\frac{\pi}{3}+1}$

$\small f'(x)=\frac{(\pi-1)(1-e)}{3}x^{\frac{\pi}{3}-1}$

Correct answer:

$\small f'(x)=\frac{\pi(1-e)}{3}x^{\frac{\pi}{3}-1}$

Explanation:

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

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

with $a=\frac{\pi}{3}$ to get

$\small f'(x)=(1-e)\cdot\frac{\pi}{3}x^{\frac{\pi}{3}-1}=\frac{\pi(1-e)}{3}x^{\frac{\pi}{3}-1}$

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Example Question #78 : Calculus Review

Find the derivative $\small f'(x)$ of the function $\small f(x)=25x^{2\pi+e^2}$

Possible Answers:

$\small f'(x)=25\cdot(2\pi+e^2)x^{2\pi+e^2}$

$\small f'(x)=25\cdot(2\pi+e^2-1)x^{2\pi+e^2}$

$\small f'(x)=25\cdot(2\pi+e^2)x^{2\pi+e^2-1}$

$\small f'(x)=\frac{25}{2\pi+e^2+1}x^{2\pi+e^2+1}$

Correct answer:

$\small f'(x)=25\cdot(2\pi+e^2)x^{2\pi+e^2-1}$

Explanation:

We can find the derivative $\small f'(x)$ of the function $\small f(x)=25x^{2\pi+e^2}$ by using the power rule for derivatives:

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

with $a=2\pi+e^2$ to get

$\small f'(x)=25\cdot(2\pi+e^2)x^{2\pi+e^2-1}$

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Example Question #71 : Calculus Review

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

Possible Answers:

$\small f'(x)=\left ( \frac{\pi}{e}\right )^2 x^{\frac{\pi}{e}+1}$

$\small f'(x)=\left ( \frac{\pi}{e}\right )\left ( \frac{\pi}{e}-1\right ) x^{\frac{\pi-e}{e}}$

$\small f'(x)=\left ( \frac{\pi}{e}-1\right)^2 x^{\frac{\pi-e}{e}}$

$\small f'(x)=\left ( \frac{\pi}{e}\right )^2 x^{\frac{\pi-e}{e}}$

Correct answer:

$\small f'(x)=\left ( \frac{\pi}{e}\right )^2 x^{\frac{\pi-e}{e}}$

Explanation:

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

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

with $a=\frac{\pi}{e}$ to get

$\small f'(x)=\frac{\pi}{e}\cdot\frac{\pi}{e}x^{\frac{\pi}{e}-1}=\left ( \frac{\pi}{e}\right )^2 x^{\frac{\pi-e}{e}}$

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Calculus 3 : Calculus Review

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #71 : Derivatives

Example Question #71 : Derivatives

Example Question #72 : Calculus Review

Example Question #73 : Calculus Review

Example Question #74 : Calculus Review

Example Question #71 : Derivatives

Example Question #76 : Calculus Review

Example Question #77 : Calculus Review

Example Question #78 : Calculus Review

Example Question #71 : Calculus Review

All Calculus 3 Resources

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