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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 #416 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&-10tan(20tan(6x))\end{align*}$

Possible Answers:

$120tan(6x)^{2} + 120$

$- 10tan(20tan(6x))^{2} - 10$

$-10tan(120tan(6x)^{2} + 120)$

$-10\cdot (120tan(6x)^{2} + 120)\cdot (tan(20tan(6x))^{2} + 1)$

Correct answer:

$-10\cdot (120tan(6x)^{2} + 120)\cdot (tan(20tan(6x))^{2} + 1)$

Explanation:

$\begin{align*}&\text{Our function, }-10tan(20tan(6x))\text{, is one function nested within another.}\\&\text{Note the }20tan(6x)\text{ inside of the }-10tan()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[tan(u)]=\frac{du}{cos^2(u)}=(tan^2(u)+1)du\\&u=20tan(6x)\\&du=120tan(6x)^{2} + 120\\&f'(x)=(120tan(6x)^{2} + 120)(- 10tan(20tan(6x))^{2} - 10)=-10\cdot (120tan(6x)^{2} + 120)\cdot (tan(20tan(6x))^{2} + 1)\end{align*}$

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Example Question #417 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-17e^{(-19\cdot 5^{(4x)})}\end{align*}$

Possible Answers:

$-76\cdot 5^{(4x)}ln(5)$

$-17e^{(-76\cdot 5^{(4x)}ln(5))}$

$-17e^{(-19\cdot 5^{(4x)})}$

$1292\cdot 5^{(4x)}e^{(-19\cdot 5^{(4x)})}ln(5)$

Correct answer:

$1292\cdot 5^{(4x)}e^{(-19\cdot 5^{(4x)})}ln(5)$

Explanation:

$\begin{align*}&\text{The function }y=-17e^{(-19\cdot 5^{(4x)})}\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }-19\cdot 5^{(4x)}\text{ is inside of the }-17e^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[e^u]=e^udu\\&u=-19\cdot 5^{(4x)}\\&du=-76\cdot 5^{(4x)}ln(5)\\&\frac{dy}{dx}=(-76\cdot 5^{(4x)}ln(5))(-17e^{(-19\cdot 5^{(4x)})})=1292\cdot 5^{(4x)}e^{(-19\cdot 5^{(4x)})}ln(5)\end{align*}$

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Example Question #411 : Derivative Review

$\begin{align*}&\text{Calculate the derivative of the function}\\&4^{(15cos(4x))}\end{align*}$

Possible Answers:

$-60\cdot 4^{(15cos(4x))}sin(4x)ln(4)$

$\frac{1}{4^{(60sin(4x))}}$

$\frac{81}{4^{(60sin(4x))}}$

-60sin(4x)

Correct answer:

$-60\cdot 4^{(15cos(4x))}sin(4x)ln(4)$

Explanation:

$\begin{align*}&\text{Our function, }4^{(15cos(4x))}\text{, is one function nested within another.}\\&\text{Note the }15cos(4x)\text{ inside of the }1\cdot4^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[a^u]=a^uduln(a)\\&u=15cos(4x)\\&du=-60sin(4x)\\&f'(x)=(-60sin(4x))(4^{(15cos(4x))}ln(4))=-60\cdot 4^{(15cos(4x))}sin(4x)ln(4)\end{align*}$

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Example Question #419 : Derivative Review

$\begin{align*}&\text{Find the derivative of the function }\\&f(x)=\frac{1}{5^{(17x^{2})}}\end{align*}$

Possible Answers:

$\frac{1}{5^{(34x)}}$

$-\frac{(68xln(5))}{5^{(17x^{2})}}$

$-\frac{(34xln(5))}{5^{(17x^{2})}}$

-34x

Correct answer:

$-\frac{(34xln(5))}{5^{(17x^{2})}}$

Explanation:

$\begin{align*}&\text{The function }f(x)=\frac{1}{5^{(17x^{2})}}\text{, can be seen as a function nested within another.}\\&\text{Notice the }-17x^{2}\text{ within the }1\cdot5^{()}\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&u=-17x^{2}\\&du=-34x\\&f'(x)=(-34x)(\frac{ln(5)}{5^{(17x^{2})}})=-\frac{(34xln(5))}{5^{(17x^{2})}}\end{align*}$

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Example Question #420 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-13ln(16cos(3x))\end{align*}$

Possible Answers:

$\frac{(39sin(3x))}{cos(3x)}$

-48sin(3x)

$-\frac{13}{(16cos(3x))}$

-13ln(-48sin(3x))

Correct answer:

$\frac{(39sin(3x))}{cos(3x)}$

Explanation:

$\begin{align*}&\text{The function }y=-13ln(16cos(3x))\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }16cos(3x)\text{ is inside of the }-13ln()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&d[ln(u)]=\frac{du}{u}\\&u=16cos(3x)\\&du=-48sin(3x)\\&\frac{dy}{dx}=(-48sin(3x))(-\frac{13}{(16cos(3x))})=\frac{(39sin(3x))}{cos(3x)}\end{align*}$

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Example Question #221 : First And Second Derivatives Of Functions

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

Possible Answers:

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

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

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

$-14e^{(x)}$

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 #222 : First And Second Derivatives Of Functions

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

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 #223 : First And Second Derivatives Of Functions

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

Possible Answers:

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

$\frac{3}{x}$

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

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

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 #224 : First And Second Derivatives Of Functions

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

Possible Answers:

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

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

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

10cos(7tan(2x))

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 #225 : First And Second Derivatives Of Functions

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

Possible Answers:

16sin(11sin(2x))

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

-22cos(2x)

16cos(22cos(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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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #416 : Derivative Review

Example Question #417 : Derivative Review

Example Question #411 : Derivative Review

Example Question #419 : Derivative Review

Example Question #420 : Derivative Review

Example Question #221 : First And Second Derivatives Of Functions

Example Question #222 : First And Second Derivatives Of Functions

Example Question #223 : First And Second Derivatives Of Functions

Example Question #224 : First And Second Derivatives Of Functions

Example Question #225 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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