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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 #1521 : Calculus Ii

Find the derivative of the function:

$f(x)=\sqrt{\tan(2x)}$

Possible Answers:

$\frac{2\sec^2(2x)}{\sqrt{\tan(2x)}}$

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

$\frac{\sec^2(2x)}{2\sqrt{\tan(2x)}}$

$\frac{\sec^2(2x)}{\sqrt{\tan(2x)}}$

Correct answer:

$\frac{\sec^2(2x)}{\sqrt{\tan(2x)}}$

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

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Example Question #397 : Derivative Review

Use implicit differentiation to find $\frac{dy}{dx}$ if y=cos(x+y)

Possible Answers:

-sin(x+y)

$\frac{-cos(x+y)}{1+cos(x+y)}$

sin(x)

$\frac{-sin(x+y)}{1+sin(x+y)}$

sin(x+y)

Correct answer:

$\frac{-sin(x+y)}{1+sin(x+y)}$

Explanation:

$\frac{dy}{dx}=-sin(x+y)(1+\frac{dy}{dx})$

$\frac{dy}{dx}=-sin(x+y)-\frac{dy}{dx}sin(x+y)$

$\frac{dy}{dx}+\frac{dy}{dx}sin(x+y)=-sin(x+y)$

$\frac{dy}{dx}(1+sin(x+y))=-sin(x+y)$

$\frac{dy}{dx}=\frac{-sin(x+y)}{(1+sin(x+y))}$

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Example Question #398 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&-\frac{7}{2^{(4tan(4x))}}\end{align*}$

Possible Answers:

$\frac{(7ln(2)\cdot (16tan(4x)^{2} + 16))}{2^{(4tan(4x))}}$

$-\frac{7}{2^{(16tan(4x)^{2} + 16)}}$

$-\frac{4370}{(9002\cdot 2^{(4tan(4x))})}$

$- 16tan(4x)^{2} - 16$

Correct answer:

$\frac{(7ln(2)\cdot (16tan(4x)^{2} + 16))}{2^{(4tan(4x))}}$

Explanation:

$\begin{align*}&\text{Our function, }-\frac{7}{2^{(4tan(4x))}}\text{, is one function nested within another.}\\&\text{Note the }-4tan(4x)\text{ inside of the }-7\cdot2^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[a^u]=a^uduln(a)\\&u=-4tan(4x)\\&du=- 16tan(4x)^{2} - 16\\&f'(x)=(- 16tan(4x)^{2} - 16)(-\frac{(7ln(2))}{2^{(4tan(4x))}})=\frac{(7ln(2)\cdot (16tan(4x)^{2} + 16))}{2^{(4tan(4x))}}\end{align*}$

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Example Question #399 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&tan(6ln(5x))\end{align*}$

Possible Answers:

$-\frac{6}{x}$

$- tan(6ln(5x))^{2} - 1$

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

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

Correct answer:

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

Explanation:

$\begin{align*}&\text{Our function, }tan(6ln(5x))\text{, is one function nested within another.}\\&\text{Note the }-6ln(5x)\text{ inside of the }-1tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[ln(u)]=\frac{du}{u}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=-6ln(5x)\\&du=-\frac{6}{x}\\&f'(x)=(-\frac{6}{x})(- tan(6ln(5x))^{2} - 1)=\frac{(6\cdot (tan(6ln(5x))^{2} + 1))}{x}\end{align*}$

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Example Question #191 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=5\cdot 3^{(3x^{2})}\end{align*}$

Possible Answers:

$30\cdot 3^{(3x^{2})}xln(3)$

$\frac{(12369274734675865\cdot 3^{(3x^{2})})}{2251799813685248}$

$5\cdot 3^{(6x)}$

Correct answer:

$30\cdot 3^{(3x^{2})}xln(3)$

Explanation:

$\begin{align*}&\text{The function }f(x)=5\cdot 3^{(3x^{2})}\text{, can be seen as a function nested within another.}\\&\text{Notice the }3x^{2}\text{ within the }5\cdot3^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&u=3x^{2}\\&du=6x\\&f'(x)=(6x)(5\cdot 3^{(3x^{2})}ln(3))=30\cdot 3^{(3x^{2})}xln(3)\end{align*}$

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Example Question #201 : First And Second Derivatives Of Functions

$\begin{align*}&\text{Calculate the derivative of the function}\\&17cos(16sin(6x))\end{align*}$

Possible Answers:

-17sin(16sin(6x))

17cos(96cos(6x))

96cos(6x)

-1632sin(16sin(6x))cos(6x)

Correct answer:

-1632sin(16sin(6x))cos(6x)

Explanation:

$\begin{align*}&\text{Our function, }17cos(16sin(6x))\text{, is one function nested within another.}\\&\text{Note the }16sin(6x)\text{ inside of the }17cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&u=16sin(6x)\\&du=96cos(6x)\\&f'(x)=(96cos(6x))(-17sin(16sin(6x)))=-1632sin(16sin(6x))cos(6x)\end{align*}$

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Example Question #1521 : Calculus Ii

$\begin{align*}&\text{Calculate the derivative of the function}\\&-17sin(15sin(x))\end{align*}$

Possible Answers:

-17cos(15sin(x))

-17sin(15cos(x))

15cos(x)

-255cos(15sin(x))cos(x)

Correct answer:

-255cos(15sin(x))cos(x)

Explanation:

$\begin{align*}&\text{Our function, }-17sin(15sin(x))\text{, is one function nested within another.}\\&\text{Note the }15sin(x)\text{ inside of the }-17sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&u=15sin(x)\\&du=15cos(x)\\&f'(x)=(15cos(x))(-17cos(15sin(x)))=-255cos(15sin(x))cos(x)\end{align*}$

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Example Question #402 : Derivatives

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=20e^{(5tan(4x))}\end{align*}$

Possible Answers:

$20tan(4x)^{2} + 20$

$20e^{(5tan(4x))}\cdot (20tan(4x)^{2} + 20)$

$20e^{(5tan(4x))}$

$20e^{(20tan(4x)^{2} + 20)}$

Correct answer:

$20e^{(5tan(4x))}\cdot (20tan(4x)^{2} + 20)$

Explanation:

$\begin{align*}&\text{The function }f(x)=20e^{(5tan(4x))}\text{, can be seen as a function nested within another.}\\&\text{Notice the }5tan(4x)\text{ within the }20e^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[e^u]=e^udu\\&u=5tan(4x)\\&du=20tan(4x)^{2} + 20\\&f'(x)=(20tan(4x)^{2} + 20)(20e^{(5tan(4x))})=20e^{(5tan(4x))}\cdot (20tan(4x)^{2} + 20)\end{align*}$

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Example Question #402 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=19tan(17cos(5x))\end{align*}$

Possible Answers:

-19tan(85sin(5x))

$-1615sin(5x)\cdot (tan(17cos(5x))^{2} + 1)$

$19tan(17cos(5x))^{2} + 19$

-85sin(5x)

Correct answer:

$-1615sin(5x)\cdot (tan(17cos(5x))^{2} + 1)$

Explanation:

$\begin{align*}&\text{The function }f(x)=19tan(17cos(5x))\text{, can be seen as a function nested within another.}\\&\text{Notice the }17cos(5x)\text{ within the }19tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=17cos(5x)\\&du=-85sin(5x)\\&f'(x)=(-85sin(5x))(19tan(17cos(5x))^{2} + 19)=-1615sin(5x)\cdot (tan(17cos(5x))^{2} + 1)\end{align*}$

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Example Question #1522 : Calculus Ii

$\begin{align*}&\text{Calculate the derivative of the function}\\&-6sin(4cos(3x))\end{align*}$

Possible Answers:

6sin(12sin(3x))

6cos(4cos(3x))

72sin(3x)cos(4cos(3x))

Correct answer:

72sin(3x)cos(4cos(3x))

Explanation:

$\begin{align*}&\text{Our function, }-6sin(4cos(3x))\text{, is one function nested within another.}\\&\text{Note the }-4cos(3x)\text{ inside of the }6sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&u=-4cos(3x)\\&du=12sin(3x)\\&f'(x)=(12sin(3x))(6cos(4cos(3x)))=72sin(3x)cos(4cos(3x))\end{align*}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1521 : Calculus Ii

Example Question #397 : Derivative Review

Example Question #398 : Derivative Review

Example Question #399 : Derivative Review

Example Question #191 : First And Second Derivatives Of Functions

Example Question #201 : First And Second Derivatives Of Functions

Example Question #1521 : Calculus Ii

Example Question #402 : Derivatives

Example Question #402 : Derivative Review

Example Question #1522 : Calculus Ii

All Calculus 2 Resources

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