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Calculus 2 : Derivative Review

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Example Questions

Example Question #381 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&12cos(14e^{(x)})\end{align*}$

Possible Answers:

$-168e^{(x)}sin(14e^{(x)})$

$12cos(14e^{(x)})$

$14e^{(x)}$

$-12sin(14e^{(x)})$

Correct answer:

$-168e^{(x)}sin(14e^{(x)})$

Explanation:

$\begin{align*}&\text{Our function, }12cos(14e^{(x)})\text{, is one function nested within another.}\\&\text{Note the }14e^{(x)}\text{ inside of the }12cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[e^u]=e^udu\\&d[cos(u)]=-sin(u)du\\&u=14e^{(x)}\\&du=14e^{(x)}\\&f'(x)=14e^{(x)}(-12sin(14e^{(x)}))=-168e^{(x)}sin(14e^{(x)})\end{align*}$

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Example Question #382 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-9sin(15cos(3x))\end{align*}$

Possible Answers:

405sin(3x)cos(15cos(3x))

9sin(45sin(3x))

45sin(3x)

9cos(15cos(3x))

Correct answer:

405sin(3x)cos(15cos(3x))

Explanation:

$\begin{align*}&\text{The function }f(x)=-9sin(15cos(3x))\text{, can be seen as a function nested within another.}\\&\text{Notice the }-15cos(3x)\text{ within the }9sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[sin(u)]=cos(u)du\\&u=-15cos(3x)\\&du=45sin(3x)\\&f'(x)=45sin(3x)(9cos(15cos(3x)))=405sin(3x)cos(15cos(3x))\end{align*}$

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Example Question #383 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&13ln(10ln(x))\end{align*}$

Possible Answers:

$\frac{10}{x}$

$13ln(\frac{10}{x})$

$\frac{13}{(10ln(x))}$

$\frac{13}{(xln(x))}$

Correct answer:

$\frac{13}{(xln(x))}$

Explanation:

$\begin{align*}&\text{Our function, }13ln(10ln(x))\text{, is one function nested within another.}\\&\text{Note the }10ln(x)\text{ inside of the }13ln()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[ln(u)]=\frac{du}{u}\\&u=10ln(x)\\&du=\frac{10}{x}\\&f'(x)=\frac{10}{x}(\frac{13}{(10ln(x))})=\frac{13}{(xln(x))}\end{align*}$

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Example Question #384 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-11tan(12ln(3x))\end{align*}$

Possible Answers:

$-\frac{12}{x}$

$11tan(12ln(3x))^{2} + 11$

$-\frac{(132\cdot (tan(12ln(3x))^{2} + 1))}{x}$

$-11tan(\frac{12}{x})$

Correct answer:

$-\frac{(132\cdot (tan(12ln(3x))^{2} + 1))}{x}$

Explanation:

$\begin{align*}&\text{The function }y=-11tan(12ln(3x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-12ln(3x)\text{ is inside of the }11tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[ln(u)]=\frac{du}{u}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=-12ln(3x)\\&du=-\frac{12}{x}\\&\frac{dy}{dx}=-\frac{12}{x}(11tan(12ln(3x))^{2} + 11)=-\frac{(132\cdot (tan(12ln(3x))^{2} + 1))}{x}\end{align*}$

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Example Question #384 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-5sin(9tan(3x))\end{align*}$

Possible Answers:

-5cos(9tan(3x))

$-5sin(27tan(3x)^{2} + 27)$

$-5cos(9tan(3x))\cdot (27tan(3x)^{2} + 27)$

$27tan(3x)^{2} + 27$

Correct answer:

$-5cos(9tan(3x))\cdot (27tan(3x)^{2} + 27)$

Explanation:

$\begin{align*}&\text{The function }y=-5sin(9tan(3x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }9tan(3x)\text{ is inside of the }-5sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&u=9tan(3x)\\&du=27tan(3x)^{2} + 27\\&\frac{dy}{dx}=27tan(3x)^{2} + 27(-5cos(9tan(3x)))=-5cos(9tan(3x))\cdot (27tan(3x)^{2} + 27)\end{align*}$

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Example Question #183 : First And Second Derivatives Of Functions

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

Possible Answers:

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

100cos(5x)

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

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

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 #384 : Derivative Review

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

Possible Answers:

-6tan(3cos(x))

3cos(x)

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

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

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 #385 : 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:

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

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

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

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

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 #386 : Derivative Review

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

Possible Answers:

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

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

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

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

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 #387 : Derivative Review

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

Possible Answers:

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

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

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

$-\frac{6}{x}$

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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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #381 : Derivative Review

Example Question #382 : Derivative Review

Example Question #383 : Derivative Review

Example Question #384 : Derivative Review

Example Question #384 : Derivative Review

Example Question #183 : First And Second Derivatives Of Functions

Example Question #384 : Derivative Review

Example Question #385 : Derivative Review

Example Question #386 : Derivative Review

Example Question #387 : Derivative Review

All Calculus 2 Resources

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