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Calculus 2 : Derivatives

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Example Questions

Example Question #201 : First And Second Derivatives Of Functions

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

Possible Answers:

$\displaystyle 96cos(6x)$

$\displaystyle 17cos(96cos(6x))$

$\displaystyle -17sin(16sin(6x))$

$\displaystyle -1632sin(16sin(6x))cos(6x)$

Correct answer:

$\displaystyle -1632sin(16sin(6x))cos(6x)$

Explanation:

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

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Example Question #401 : Derivatives

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

Possible Answers:

$\displaystyle -17cos(15sin(x))$

$\displaystyle -17sin(15cos(x))$

$\displaystyle -255cos(15sin(x))cos(x)$

$\displaystyle 15cos(x)$

Correct answer:

$\displaystyle -255cos(15sin(x))cos(x)$

Explanation:

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

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Example Question #403 : Derivative Review

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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Example Question #404 : Derivative Review

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

Possible Answers:

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

$\displaystyle -85sin(5x)$

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

$\displaystyle -19tan(85sin(5x))$

Correct answer:

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

Explanation:

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

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Example Question #402 : Derivatives

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

Possible Answers:

$\displaystyle 6cos(4cos(3x))$

$\displaystyle 72sin(3x)cos(4cos(3x))$

$\displaystyle 6sin(12sin(3x))$

Correct answer:

$\displaystyle 72sin(3x)cos(4cos(3x))$

Explanation:

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

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Example Question #403 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-cos(10tan(5x))\end{align*}$ $\displaystyle \begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-cos(10tan(5x))\end{align*}$

Possible Answers:

$\displaystyle sin(10tan(5x))$

$-cos(50tan(5x)^{2} + 50)$ $\displaystyle -cos(50tan(5x)^{2} + 50)$

$sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)$ $\displaystyle sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)$

$50tan(5x)^{2} + 50$ $\displaystyle 50tan(5x)^{2} + 50$

Correct answer:

$sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)$ $\displaystyle sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)$

Explanation:

$\begin{align*}&\text{The function }y=-cos(10tan(5x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }10tan(5x)\text{ is inside of the }-1cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&u=10tan(5x)\\&du=50tan(5x)^{2} + 50\\&\frac{dy}{dx}=(50tan(5x)^{2} + 50)(sin(10tan(5x)))=sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)\end{align*}$ $\displaystyle \begin{align*}&\text{The function }y=-cos(10tan(5x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }10tan(5x)\text{ is inside of the }-1cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[cos(u)]=-sin(u)du\\&u=10tan(5x)\\&du=50tan(5x)^{2} + 50\\&\frac{dy}{dx}=(50tan(5x)^{2} + 50)(sin(10tan(5x)))=sin(10tan(5x))\cdot (50tan(5x)^{2} + 50)\end{align*}$

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Example Question #403 : Derivatives

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=4cos(7sin(x))\end{align*}$ $\displaystyle \begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=4cos(7sin(x))\end{align*}$

Possible Answers:

$\displaystyle 4sin(7sin(x))$

$\displaystyle 4cos(7cos(x))$

$\displaystyle -28sin(7sin(x))cos(x)$

$\displaystyle -7cos(x)$

Correct answer:

$\displaystyle -28sin(7sin(x))cos(x)$

Explanation:

$\begin{align*}&\text{The function }y=4cos(7sin(x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-7sin(x)\text{ is inside of the }4cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&u=-7sin(x)\\&du=-7cos(x)\\&\frac{dy}{dx}=(-7cos(x))(4sin(7sin(x)))=-28sin(7sin(x))cos(x)\end{align*}$ $\displaystyle \begin{align*}&\text{The function }y=4cos(7sin(x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-7sin(x)\text{ is inside of the }4cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&u=-7sin(x)\\&du=-7cos(x)\\&\frac{dy}{dx}=(-7cos(x))(4sin(7sin(x)))=-28sin(7sin(x))cos(x)\end{align*}$

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Example Question #405 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&10cos(2\cdot 3^x)\end{align*}$ $\displaystyle \begin{align*}&\text{Calculate the derivative of the function}\\&10cos(2\cdot 3^x)\end{align*}$

Possible Answers:

$10sin(2\cdot 3^x)$ $\displaystyle 10sin(2\cdot 3^x)$

$-2\cdot 3^xln(3)$ $\displaystyle -2\cdot 3^xln(3)$

$10cos(2\cdot 3^xln(3))$ $\displaystyle 10cos(2\cdot 3^xln(3))$

$-20\cdot 3^xln(3)sin(2\cdot 3^x)$ $\displaystyle -20\cdot 3^xln(3)sin(2\cdot 3^x)$

Correct answer:

$-20\cdot 3^xln(3)sin(2\cdot 3^x)$ $\displaystyle -20\cdot 3^xln(3)sin(2\cdot 3^x)$

Explanation:

$\begin{align*}&\text{Our function, }10cos(2\cdot 3^x)\text{, is one function nested within another.}\\&\text{Note the }-2\cdot 3^x\text{ inside of the }10cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&u=-2\cdot 3^x\\&du=-2\cdot 3^xln(3)\\&f'(x)=(-2\cdot 3^xln(3))(10sin(2\cdot 3^x))=-20\cdot 3^xln(3)sin(2\cdot 3^x)\end{align*}$ $\displaystyle \begin{align*}&\text{Our function, }10cos(2\cdot 3^x)\text{, is one function nested within another.}\\&\text{Note the }-2\cdot 3^x\text{ inside of the }10cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[cos(u)]=-sin(u)du\\&u=-2\cdot 3^x\\&du=-2\cdot 3^xln(3)\\&f'(x)=(-2\cdot 3^xln(3))(10sin(2\cdot 3^x))=-20\cdot 3^xln(3)sin(2\cdot 3^x)\end{align*}$

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Example Question #409 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=5sin(3e^{(6x)})\end{align*}$ $\displaystyle \begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=5sin(3e^{(6x)})\end{align*}$

Possible Answers:

$18e^{(6x)}$ $\displaystyle 18e^{(6x)}$

$5cos(3e^{(6x)})$ $\displaystyle 5cos(3e^{(6x)})$

$5sin(18e^{(6x)})$ $\displaystyle 5sin(18e^{(6x)})$

$90e^{(6x)}cos(3e^{(6x)})$ $\displaystyle 90e^{(6x)}cos(3e^{(6x)})$

Correct answer:

$90e^{(6x)}cos(3e^{(6x)})$ $\displaystyle 90e^{(6x)}cos(3e^{(6x)})$

Explanation:

$\begin{align*}&\text{The function }y=5sin(3e^{(6x)})\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }3e^{(6x)}\text{ is inside of the }5sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&u=3e^{(6x)}\\&du=18e^{(6x)}\\&\frac{dy}{dx}=(18e^{(6x)})(5cos(3e^{(6x)}))=90e^{(6x)}cos(3e^{(6x)})\end{align*}$ $\displaystyle \begin{align*}&\text{The function }y=5sin(3e^{(6x)})\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }3e^{(6x)}\text{ is inside of the }5sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[e^u]=e^udu\\&d[sin(u)]=cos(u)du\\&u=3e^{(6x)}\\&du=18e^{(6x)}\\&\frac{dy}{dx}=(18e^{(6x)})(5cos(3e^{(6x)}))=90e^{(6x)}cos(3e^{(6x)})\end{align*}$

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Example Question #404 : Derivatives

The Laplace Transform is an integral transform that converts functions from the time domain $\displaystyle t$ to the complex frequency domain $\displaystyle s$ . The transformation of a function f(t) $\displaystyle f(t)$ into its Laplace Transform F(s) $\displaystyle F(s)$ is given by:

$F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$ $\displaystyle F(s)= \int_{0}^{\infty}e^{-st}*f(t) dt$

Where s=a+bi $\displaystyle s=a+bi$ , where $\displaystyle a$ and $\displaystyle b$ are constants and $\displaystyle i$ is the imaginary number.

Determine the value of $\frac{\mathrm{d} }{\mathrm{d} s}F(s)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} s}F(s)$ .

Possible Answers:

$\displaystyle f'(t)*F(s)$

tF(s) $\displaystyle tF(s)$

-sF(s) $\displaystyle -sF(s)$

-tF(s) $\displaystyle -tF(s)$

Correct answer:

-tF(s) $\displaystyle -tF(s)$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} s}F(s)= \frac{d}{ds}\int_{0}^{\infty}e^{-st}*f(t) dt=-t*[\int_{0}^{\infty}e^{-st}*f(t) dt]=-tF(s)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} s}F(s)= \frac{d}{ds}\int_{0}^{\infty}e^{-st}*f(t) dt=-t*[\int_{0}^{\infty}e^{-st}*f(t) dt]=-tF(s)$

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #201 : First And Second Derivatives Of Functions

Example Question #401 : Derivatives

Example Question #403 : Derivative Review

Example Question #404 : Derivative Review

Example Question #402 : Derivatives

Example Question #403 : Derivative Review

Example Question #403 : Derivatives

Example Question #405 : Derivative Review

Example Question #409 : Derivative Review

Example Question #404 : Derivatives

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