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Calculus 2 : Calculus II

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #312 : Derivatives

Find the derivative of the function:

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

Possible Answers:

$\sec^2(e^{x^2})$

$2xe^{x^2}\sec^2(e^{x^2})$

$2xe^{x^2}\sec(e^{x^2})$

$e^{x^2}\sec^2(e^{x^2})$

Correct answer:

$2xe^{x^2}\sec^2(e^{x^2})$

Explanation:

The derivative of the function is equal to the following:

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x}e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$

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Example Question #313 : Derivatives

Find the first derivative of the function:

$f(x)=e^{\cos(x)}+e^{5x}$

Possible Answers:

$-\sin(x)e^{cos(x)}+5e^{5x}+C$

$-5\sin(x)e^{\cos(x)}$

$-\sin(x)e^{cos(x)}+5e^{5x}$

$\sin(x)e^{cos(x)}+5e^{5x}$

Correct answer:

$-\sin(x)e^{cos(x)}+5e^{5x}$

Explanation:

The first derivative of the function is equal to:

$f'(x)=-\sin(x)e^{cos(x)}+5e^{5x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #313 : Derivative Review

Find the second derivative of the function:

$g(x)=x^{\frac{1}{2}}\cos(x)$

Possible Answers:

$\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})$

$\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})-\sin(x)(x^{\frac{1}{2}})$

$\frac{1}{2}x^{-\frac{1}{2}}\cos(x)-x^{\frac{1}{2}}\sin(x)$

$\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})+\sin(x)(x^{\frac{1}{2}})$

Correct answer:

$\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})$

Explanation:

The first derivative of the function is equal to

$g'(x)=\frac{1}{2}x^{-\frac{1}{2}}\cos(x)-x^{\frac{1}{2}}\sin(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

The second derivative of the function is equal to

$g''(x)=\cos(x)(-\frac{1}{4}x^{-\frac{3}{2}}-x^{\frac{1}{2}})$

and was found using the rules above, along with

$\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=\cos(x)$

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Example Question #314 : Derivatives

Find the derivative of the following function:

$f(x)=\ln(x^4e^x)$

Possible Answers:

$\frac{4}{x}+1$

$\frac{1}{x^4}$

$\frac{4}{x}$

$\frac{1}{x}+1$

$\frac{1}{x^4}+1$

Correct answer:

$\frac{4}{x}+1$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\ln(ab)=\ln(a)+\ln(b)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #117 : First And Second Derivatives Of Functions

What is the derivative of $f(x)=4x^{\frac{3}{4}}-4x+1$ ?

Possible Answers:

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

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

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

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}-4$

$f{}'(x)=\frac{6}{x^{\frac{1}{4}}}-4$

Correct answer:

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}-4$

Explanation:

To take the derivative, remember to multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=4(\frac{3}{4})x^{-\frac{1}{4}}-4$

Simplify and make the negative exponent positive by putting it on the denominator:

$f{}'(x)=\frac{3}{x^{\frac{1}{4}}}-4$ .

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Example Question #321 : Derivatives

Find the derivative of f(x)=-3x^3-9x^2+4x-1 .

Possible Answers:

$f{}'(x)=-27x^2+18x+4$

$f{}'(x)=-27x^2-18x+4$

$f{}'(x)=-27x^2-18x$

$f{}'(x)=27x^2-18x+4$

$f{}'(x)=-27x^2-18x+4x$

Correct answer:

$f{}'(x)=-27x^2-18x+4$

Explanation:

To take the derivative, remember to multiply the exponent of an x term by the coefficient in front of the x term and then subtract one from the exponent:

$f{}'(x)=-27x^2-18x+4$ .

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Example Question #122 : First And Second Derivatives Of Functions

What is the acceleration function if x(t)=-16t^2+4t-1?

Possible Answers:

a(t)=-32t^2

a(t)=-32t-1

a(t)=-32t

a(t)=-32

a(t)=-3

Correct answer:

a(t)=-32

Explanation:

Recall that the acceleration function is the second derivative of the position function. First, find the velocity function since that is the first derivative. To take the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

v(t)=-32t+4

Now, take the derivative of the velocity function to get acceleration:

a(t)=-32

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Example Question #123 : First And Second Derivatives Of Functions

Find the first derivative of the function:

$f(x)=\tan(2xe^x)$

Possible Answers:

$\sec^2(2xe^x)$

$\sec^2(2xe^x)[2e^x+2xe^x]$

$\sec^2(2xe^x)[2e^x+xe^x]$

$\sec^2(2xe^x)[4e^x]$

Correct answer:

$\sec^2(2xe^x)[2e^x+2xe^x]$

Explanation:

The first derivative of the function is equal to

$f'(x)=\sec^2(2xe^x)[2e^x+2xe^x]$

and was found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #124 : First And Second Derivatives Of Functions

Find the derivative of the following:

$f(x)=\frac{\cos(2x)}{e^{2x}}$

Possible Answers:

$\frac{\sin(2x)-\cos(2x)}{e^{2x}}$

$\frac{-2(\sin(2x)-\cos(2x))}{e^{2x}}$

$-2(\sin(2x)-\cos(2x))$

$\frac{-2(\sin(2x)+\cos(2x))}{e^{2x}}$

Correct answer:

$\frac{-2(\sin(2x)-\cos(2x))}{e^{2x}}$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{-2(\sin(2x)-\cos(2x))}{e^{2x}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #122 : First And Second Derivatives Of Functions

Find the first derivative of the following function:

$f(x)=\frac{x^4}{\cos(x)}+\csc(3x)$

Possible Answers:

$\frac{4x^3\cos(x)-x^4\sin(x)}{\cos^2(x)}-\csc(3x)\cot(3x)$

$\frac{4x^3\cos(x)-x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

$\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

$f'(x)=\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}+3\csc(3x)\cot(3x)$

Correct answer:

$\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

Explanation:

The first derivative of the function is equal to

$f'(x)=\frac{4x^3\cos(x)+x^4\sin(x)}{\cos^2(x)}-3\csc(3x)\cot(3x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} \csc(x)=-\csc(x)\cot(x)$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #312 : Derivatives

Example Question #313 : Derivatives

Example Question #313 : Derivative Review

Example Question #314 : Derivatives

Example Question #117 : First And Second Derivatives Of Functions

Example Question #321 : Derivatives

Example Question #122 : First And Second Derivatives Of Functions

Example Question #123 : First And Second Derivatives Of Functions

Example Question #124 : First And Second Derivatives Of Functions

Example Question #122 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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