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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 #184 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Calculate the derivative of the function}\\&-17e^{(20sin(5x))}\end{align*}$

Possible Answers:

100cos(5x)

$-17e^{(20sin(5x))}$

$-1700cos(5x)e^{(20sin(5x))}$

$-17e^{(100cos(5x))}$

Correct answer:

$-1700cos(5x)e^{(20sin(5x))}$

Explanation:

$\begin{align*}&\text{Our function, }-17e^{(20sin(5x))}\text{, is one function nested within another.}\\&\text{Note the }20sin(5x)\text{ inside of the }-17e^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[e^u]=e^udu\\&u=20sin(5x)\\&du=100cos(5x)\\&f'(x)=100cos(5x)(-17e^{(20sin(5x))})=-1700cos(5x)e^{(20sin(5x))}\end{align*}$

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Example Question #386 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-6tan(3sin(x))\end{align*}$

Possible Answers:

3cos(x)

$-18cos(x)\cdot (tan(3sin(x))^{2} + 1)$

-6tan(3cos(x))

$- 6tan(3sin(x))^{2} - 6$

Correct answer:

$-18cos(x)\cdot (tan(3sin(x))^{2} + 1)$

Explanation:

$\begin{align*}&\text{The function }y=-6tan(3sin(x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }3sin(x)\text{ is inside of the }-6tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=3sin(x)\\&du=3cos(x)\\&\frac{dy}{dx}=3cos(x)(- 6tan(3sin(x))^{2} - 6)=-18cos(x)\cdot (tan(3sin(x))^{2} + 1)\end{align*}$

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Example Question #387 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=14tan(17\cdot 5^{(2x)})\end{align*}$

Possible Answers:

$14tan(34\cdot 5^{(2x)}ln(5))$

$-34\cdot 5^{(2x)}ln(5)$

$476\cdot 5^{(2x)}ln(5)\cdot (tan(17\cdot 5^{(2x)})^{2} + 1)$

$- 14tan(17\cdot 5^{(2x)})^{2} - 14$

Correct answer:

$476\cdot 5^{(2x)}ln(5)\cdot (tan(17\cdot 5^{(2x)})^{2} + 1)$

Explanation:

$\begin{align*}&\text{The function }y=14tan(17\cdot 5^{(2x)})\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-17\cdot 5^{(2x)}\text{ is inside of the }-14tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=-17\cdot 5^{(2x)}\\&du=-34\cdot 5^{(2x)}ln(5)\\&\frac{dy}{dx}=-34\cdot 5^{(2x)}ln(5)(- 14tan(17\cdot 5^{(2x)})^{2} - 14)=476\cdot 5^{(2x)}ln(5)\cdot (tan(17\cdot 5^{(2x)})^{2} + 1)\end{align*}$

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Example Question #388 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&-7cos(9\cdot 3^{(3x)})\end{align*}$

Possible Answers:

$-7sin(9\cdot 3^{(3x)})$

$-27\cdot 3^{(3x)}ln(3)$

$189\cdot 3^{(3x)}sin(9\cdot 3^{(3x)})ln(3)$

$-7cos(27\cdot 3^{(3x)}ln(3))$

Correct answer:

$189\cdot 3^{(3x)}sin(9\cdot 3^{(3x)})ln(3)$

Explanation:

$\begin{align*}&\text{Our function, }-7cos(9\cdot 3^{(3x)})\text{, is one function nested within another.}\\&\text{Note the }-9\cdot 3^{(3x)}\text{ inside of the }-7cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&u=-9\cdot 3^{(3x)}\\&du=-27\cdot 3^{(3x)}ln(3)\\&f'(x)=-27\cdot 3^{(3x)}ln(3)(-7sin(9\cdot 3^{(3x)}))=189\cdot 3^{(3x)}sin(9\cdot 3^{(3x)})ln(3)\end{align*}$

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Example Question #389 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=11tan(6ln(5x))\end{align*}$

Possible Answers:

$\frac{(66\cdot (tan(6ln(5x))^{2} + 1))}{x}$

$11tan(\frac{6}{x})$

$- 11tan(6ln(5x))^{2} - 11$

$-\frac{6}{x}$

Correct answer:

$\frac{(66\cdot (tan(6ln(5x))^{2} + 1))}{x}$

Explanation:

$\begin{align*}&\text{The function }y=11tan(6ln(5x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-6ln(5x)\text{ is inside of the }-11tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[ln(u)]=\frac{du}{u}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=-6ln(5x)\\&du=-\frac{6}{x}\\&\frac{dy}{dx}=-\frac{6}{x}(- 11tan(6ln(5x))^{2} - 11)=\frac{(66\cdot (tan(6ln(5x))^{2} + 1))}{x}\end{align*}$

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Example Question #391 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-20sin(8\cdot 3^{(5x)})\end{align*}$

Possible Answers:

$40\cdot 3^{(5x)}ln(3)$

$-20sin(40\cdot 3^{(5x)}ln(3))$

$-20cos(8\cdot 3^{(5x)})$

$-800\cdot 3^{(5x)}ln(3)cos(8\cdot 3^{(5x)})$

Correct answer:

$-800\cdot 3^{(5x)}ln(3)cos(8\cdot 3^{(5x)})$

Explanation:

$\begin{align*}&\text{The function }y=-20sin(8\cdot 3^{(5x)})\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }8\cdot 3^{(5x)}\text{ is inside of the }-20sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&u=8\cdot 3^{(5x)}\\&du=40\cdot 3^{(5x)}ln(3)\\&\frac{dy}{dx}=40\cdot 3^{(5x)}ln(3)(-20cos(8\cdot 3^{(5x)}))=-800\cdot 3^{(5x)}ln(3)cos(8\cdot 3^{(5x)})\end{align*}$

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Example Question #392 : Derivative Review

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #393 : Derivative Review

Find the second derivative of the function:

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

Possible Answers:

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

$\frac{2}{x}$

$\frac{2}{x^2}$

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

Correct answer:

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

Explanation:

The first derivative of the function is equal to

$f'(x)=\frac{2}{x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The second derivative of the function is equal to

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

and was found using a rule stated above.

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Example Question #394 : Derivative Review

Find the derivative of the function:

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

Possible Answers:

$\frac{2x-x^2}{e^x}$

$\frac{2x+x^2}{e^{x}}$

$\frac{2x+x^2}{e^{2x}}$

$\frac{2x-x^2}{e^{2x}}$

Correct answer:

$\frac{2x-x^2}{e^x}$

Explanation:

The derivative of the function is

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

and was found using the following rules:

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

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Example Question #395 : Derivative Review

Find the derivative of the function:

$f(x)=x^2\tan(x^3)$

Possible Answers:

$2x\tan(x^3)+3x^2\sec^2(x^3)$

$2x\tan(x^3)-3x^4\sec^2(x^3)$

$2x\tan(x^3)+x^4\sec^2(x^3)$

$2x\tan(x^3)+3x^4\sec^2(x^3)$

Correct answer:

$2x\tan(x^3)+3x^4\sec^2(x^3)$

Explanation:

The derivative of the function is equal to

$f'(x)=2x\tan(x^3)+3x^4\sec^2(x^3)$

and was found using the following rules:

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

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #184 : First And Second Derivatives Of Functions

Example Question #386 : Derivative Review

Example Question #387 : Derivative Review

Example Question #388 : Derivative Review

Example Question #389 : Derivative Review

Example Question #391 : Derivative Review

Example Question #392 : Derivative Review

Example Question #393 : Derivative Review

Example Question #394 : Derivative Review

Example Question #395 : Derivative Review

All Calculus 2 Resources

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