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Calculus 2 : Derivative Review

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Example Questions

Example Question #421 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=-10e^{(-14e^{(x)})}\end{align*}$

Possible Answers:

$140e^{(-14e^{(x)})}e^{(x)}$

$10e^{(-14e^{(x)})}$

$-10e^{(-14e^{(x)})}$

$-14e^{(x)}$

Correct answer:

$140e^{(-14e^{(x)})}e^{(x)}$

Explanation:

$\begin{align*}&\text{The function }f(x)=-10e^{(-14e^{(x)})}\text{, can be seen as a function nested within another.}\\&\text{Notice the }-14e^{(x)}\text{ within the }-10e^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[e^u]=e^udu\\&u=-14e^{(x)}\\&du=-14e^{(x)}\\&f'(x)=(-14e^{(x)})(-10e^{(-14e^{(x)})})=140e^{(-14e^{(x)})}e^{(x)}\end{align*}$

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Example Question #422 : Derivative Review

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

Possible Answers:

-10cos(2x)

-14cos(5sin(2x))

140cos(5sin(2x))cos(2x)

14sin(10cos(2x))

Correct answer:

140cos(5sin(2x))cos(2x)

Explanation:

$\begin{align*}&\text{The function }y=14sin(5sin(2x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-5sin(2x)\text{ is inside of the }-14sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&u=-5sin(2x)\\&du=-10cos(2x)\\&\frac{dy}{dx}=(-10cos(2x))(-14cos(5sin(2x)))=140cos(5sin(2x))cos(2x)\end{align*}$

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Example Question #423 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&-6tan(3ln(2x))\end{align*}$

Possible Answers:

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

$\frac{3}{x}$

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

$-6tan(\frac{3}{x})$

Correct answer:

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

Explanation:

$\begin{align*}&\text{Our function, }-6tan(3ln(2x))\text{, is one function nested within another.}\\&\text{Note the }3ln(2x)\text{ inside of the }-6tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[ln(u)]=\frac{du}{u}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=3ln(2x)\\&du=\frac{3}{x}\\&f'(x)=(\frac{3}{x})(- 6tan(3ln(2x))^{2} - 6)=-\frac{(18\cdot (tan(3ln(2x))^{2} + 1))}{x}\end{align*}$

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Example Question #424 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=10sin(7tan(2x))\end{align*}$

Possible Answers:

10cos(7tan(2x))

$10cos(7tan(2x))\cdot (14tan(2x)^{2} + 14)$

$10sin(14tan(2x)^{2} + 14)$

$14tan(2x)^{2} + 14$

Correct answer:

$10cos(7tan(2x))\cdot (14tan(2x)^{2} + 14)$

Explanation:

$\begin{align*}&\text{The function }y=10sin(7tan(2x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }7tan(2x)\text{ is inside of the }10sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&d[sin(u)]=cos(u)du\\&u=7tan(2x)\\&du=14tan(2x)^{2} + 14\\&\frac{dy}{dx}=(14tan(2x)^{2} + 14)(10cos(7tan(2x)))=10cos(7tan(2x))\cdot (14tan(2x)^{2} + 14)\end{align*}$

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Example Question #425 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=16cos(11sin(2x))\end{align*}$

Possible Answers:

16sin(11sin(2x))

-22cos(2x)

16cos(22cos(2x))

-352sin(11sin(2x))cos(2x)

Correct answer:

-352sin(11sin(2x))cos(2x)

Explanation:

$\begin{align*}&\text{The function }f(x)=16cos(11sin(2x))\text{, can be seen as a function nested within another.}\\&\text{Notice the }-11sin(2x)\text{ within the }16cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[sin(u)]=cos(u)du\\&d[cos(u)]=-sin(u)du\\&u=-11sin(2x)\\&du=-22cos(2x)\\&f'(x)=(-22cos(2x))(16sin(11sin(2x)))=-352sin(11sin(2x))cos(2x)\end{align*}$

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Example Question #426 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=19cos(2x^{2})\end{align*}$

Possible Answers:

$19sin(2x^{2})$

$-76xsin(2x^{2})$

19cos(4x)

-4x

Correct answer:

$-76xsin(2x^{2})$

Explanation:

$\begin{align*}&\text{The function }f(x)=19cos(2x^{2})\text{, can be seen as a function nested within another.}\\&\text{Notice the }-2x^{2}\text{ within the }19cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&u=-2x^{2}\\&du=-4x\\&f'(x)=(-4x)(19sin(2x^{2}))=-76xsin(2x^{2})\end{align*}$

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Example Question #427 : Derivative Review

$\begin{align*}&\text{Find the 2nd derivative of the function }\\&f(x)=-14e^{(16x)}\end{align*}$

Possible Answers:

$-224e^{(16x)}$

$-3584e^{(16x)}$

$50176e^{(32x)}$

$802816e^{(32x)}$

Correct answer:

$-3584e^{(16x)}$

Explanation:

$\text{To find the second derivative of a function, one need only be}\\&\text{cautious in performing the steps of derivation twice. For the}\\&\text{function we're given,}\text{we'll wish to use the following:}\\&d[e^{ax}]=ae^{ax}dx\\&\text{Our first derivative is thus:}\\&-224e^{(16x)}\\&\text{And our second is:}\\&-3584e^{(16x)}$

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Example Question #428 : Derivative Review

$\begin{align*}&\text{Take the derivative twice with respect to }x\text{ of the function}\\&y=-5x^{4}\end{align*}$

Possible Answers:

$-15x^{2}$

$-60x^{2}$

$1200x^{5}$

$400x^{6}$

Correct answer:

$-60x^{2}$

Explanation:

$\begin{align*}&\text{To find the second derivative of a function, one need only be}\\&\text{cautious in performing the steps of derivation twice. For the}\\&\text{function we're given,}\text{ we'll wish to use the following:}\\&d[x^a]=ax^{a-1}dx\\&\text{Our first derivative is thus:}\\&-20x^{3}\\&\text{And our second is:}\\&-60x^{2}\end{align*}$

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Example Question #429 : Derivative Review

$\begin{align*}&\text{Take the derivative twice with respect to }x\text{ of the function}\\&y=20ln(10x)\end{align*}$

Possible Answers:

$-\frac{20}{x^{2}}$

$\frac{400}{x^{2}}$

$-\frac{400}{x^{3}}$

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

Correct answer:

$-\frac{20}{x^{2}}$

Explanation:

$\begin{align*}&\text{To find the second derivative of a function, one need only be}\\&\text{cautious in performing the steps of derivation twice. For the}\\&\text{function we're given,}\text{ we'll wish to use the following:}\\&d[ln(ax)]=\frac{dx}{x}\\&d[x^a]=ax^{a-1}dx\\&\text{Our first derivative is thus:}\\&\frac{20}{x}\\&\text{And our second is:}\\&-\frac{20}{x^{2}}\end{align*}$

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Example Question #430 : Derivative Review

$\begin{align*}&\text{Find the 2nd derivative of the function }\\&f(x)=-16x^{18}\end{align*}$

Possible Answers:

$-4896x^{16}$

$82944x^{34}$

$1410048x^{33}$

$-272x^{16}$

Correct answer:

$-4896x^{16}$

Explanation:

$\begin{align*}&\text{To find the second derivative of a function, one need only be}\\&\text{cautious in performing the steps of derivation twice. For the}\\&\text{function we're given,}\text{ we'll wish to use the following:}\\&d[x^a]=ax^{a-1}dx\\&\text{Our first derivative is thus:}\\&-288x^{17}\\&\text{And our second is:}\\&-4896x^{16}\end{align*}$

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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #421 : Derivative Review

Example Question #422 : Derivative Review

Example Question #423 : Derivative Review

Example Question #424 : Derivative Review

Example Question #425 : Derivative Review

Example Question #426 : Derivative Review

Example Question #427 : Derivative Review

Example Question #428 : Derivative Review

Example Question #429 : Derivative Review

Example Question #430 : Derivative Review

All Calculus 2 Resources

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