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Calculus 2 : First and Second Derivatives of Functions

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Example Questions

Example Question #391 : Derivative Review

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\\&y=-20sin(8\cdot 3^{(5x)})\end{align*}$

Possible Answers:

$40\cdot 3^{(5x)}ln(3)$

$-20sin(40\cdot 3^{(5x)}ln(3))$

$-20cos(8\cdot 3^{(5x)})$

$-800\cdot 3^{(5x)}ln(3)cos(8\cdot 3^{(5x)})$

Correct answer:

$-800\cdot 3^{(5x)}ln(3)cos(8\cdot 3^{(5x)})$

Explanation:

$\begin{align*}&\text{The function }y=-20sin(8\cdot 3^{(5x)})\text{ consists of two functions, one nested in the other..}\\&\text{In particular, }8\cdot 3^{(5x)}\text{ is inside of the }-20sin()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[a^u]=a^uduln(a)\\&d[sin(u)]=cos(u)du\\&u=8\cdot 3^{(5x)}\\&du=40\cdot 3^{(5x)}ln(3)\\&\frac{dy}{dx}=40\cdot 3^{(5x)}ln(3)(-20cos(8\cdot 3^{(5x)}))=-800\cdot 3^{(5x)}ln(3)cos(8\cdot 3^{(5x)})\end{align*}$

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Example Question #392 : Derivative Review

Find the derivative of the function:

$f(x)=\sec(e^{2x})$

Possible Answers:

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

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

$-2e^{2x}\sec(e^{2x})\tan(e^{2x})$

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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Example Question #391 : Derivative Review

Find the second derivative of the function:

$f(x)=\ln(x^2)$

Possible Answers:

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

$-\frac{1}{x^2}$

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

$\frac{2}{x}$

Correct answer:

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

Explanation:

The first derivative of the function is equal to

$f'(x)=\frac{2}{x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

The second derivative of the function is equal to

$f''(x)=-\frac{2}{x^2}$

and was found using a rule stated above.

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Example Question #394 : Derivative Review

Find the derivative of the function:

$f(x)=\frac{x^2}{e^x}$

Possible Answers:

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

$\frac{2x+x^2}{e^{x}}$

$\frac{2x+x^2}{e^{2x}}$

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

Correct answer:

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

Explanation:

The derivative of the function is

$f'(x)=\frac{2x-x^2}{e^{x}}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f'(x)g(x)}{g(x)^2}$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #1511 : Calculus Ii

Find the derivative of the function:

$f(x)=x^2\tan(x^3)$

Possible Answers:

$2x\tan(x^3)-3x^4\sec^2(x^3)$

$2x\tan(x^3)+3x^2\sec^2(x^3)$

$2x\tan(x^3)+x^4\sec^2(x^3)$

$2x\tan(x^3)+3x^4\sec^2(x^3)$

Correct answer:

$2x\tan(x^3)+3x^4\sec^2(x^3)$

Explanation:

The derivative of the function is equal to

$f'(x)=2x\tan(x^3)+3x^4\sec^2(x^3)$

and was found using the following rules:

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

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Example Question #391 : Derivative Review

Find the derivative of the function:

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

Possible Answers:

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

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

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

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

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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Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #391 : Derivative Review

Example Question #392 : Derivative Review

Example Question #391 : Derivative Review

Example Question #394 : Derivative Review

Example Question #1511 : Calculus Ii

Example Question #391 : Derivative Review

Example Question #397 : Derivative Review

Example Question #398 : Derivative Review

Example Question #399 : Derivative Review

Example Question #191 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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