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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 #461 : Derivative Review

Define $f (x,y) = e ^{x^{2}+ y^{2}}$ .

Find $f _{xy} (x,y)$ .

Possible Answers:

$f _{xy} (x,y)= 4xy e ^{x^{2}+y^{2}}$

$f _{xy} (x,y)= xy e ^{x^{2}+y^{2}}$

$f _{xy} (x,y)=2 xy e ^{x^{2}+y^{2}}$

$f _{xy} (x,y)= xe ^{x^{2}+y^{2}} + y e ^{x^{2}+y^{2}}$

$f _{xy} (x,y)=2xe ^{x^{2}+y^{2}} + 2y e ^{x^{2}+y^{2}}$

Correct answer:

$f _{xy} (x,y)= 4xy e ^{x^{2}+y^{2}}$

Explanation:

First, find $f_{x} = \frac{\partial f}{\partial x}$ as follows:

$f (x,y) = e ^{x^{2}+ y^{2}} = e ^{x^{2} } e ^{y^{2} }$

$\frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} \left (e ^{x^{2} } e ^{y^{2} } \right )$

$= e ^{y^{2}} \cdot \frac{\partial }{\partial x} \left (e ^{x^{2} }} \right )$

$= e ^{y^{2}} \cdot \frac{\partial }{\partial x} x^{2} \cdot \left (e ^{x^{2} }} \right )$

$= 2x \cdot e ^{y^{2}} \cdot \left (e ^{x^{2} }} \right )$

$= 2x e ^{x^{2}} e ^{y^{2}}$

Therefore, $f_{x} (x,y ) = 2x e ^{x^{2}} e ^{y^{2}}$ .

Now, find $f_{xy} = \frac{\partial }{\partial y} f_{x}$ as follows:

$f_{x} (x,y )= 2x e ^{x^{2}} e ^{y^{2}}$

$\frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y} \left ( 2x e ^{x^{2}} e ^{y^{2}} \right )$

$=2x e ^{x^{2}} \frac{\partial }{\partial y} \left ( e ^{y^{2}} \right )$

$=2x e ^{x^{2}} \frac{\partial }{\partial y} \left ( y^{2} \right ) e ^{y^{2}}$

$=2x e ^{x^{2}} \cdot 2y \cdot e ^{y^{2}}$

$=4xy e ^{x^{2}} e ^{y^{2}} = 4xy e ^{x^{2}+y^{2}}$

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Example Question #461 : Derivatives

Define $f (x,y) = \cos (x^{2} + y^{2})$ .

Find $f _{xy} (x,y)$ .

Possible Answers:

$f _{xy} (x,y) = 4x \cos (x^{2} + y^{2}) + 4y \cos (x^{2} + y^{2})$

$f _{xy} (x,y) =- 4x \cos (x^{2} + y^{2}) + 4y \cos (x^{2} + y^{2})$

$f _{xy} (x,y) =- 4xy \cos (x^{2} + y^{2})$

$f _{xy} (x,y) = 4xy \cos (x^{2} + y^{2})$

$f _{xy} (x,y) = 4x \cos (x^{2} + y^{2}) - 4y \cos (x^{2} + y^{2})$

Correct answer:

$f _{xy} (x,y) =- 4xy \cos (x^{2} + y^{2})$

Explanation:

First, find $f_{x} = \frac{\partial f}{\partial x}$ as follows:

$f (x,y) = \cos (x^{2} + y^{2})$

$\frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} \cos (x^{2} + y^{2})$

$= \frac{\partial }{\partial x}(x^{2} + y^{2}) \cdot \left [ - \sin (x^{2} + y^{2}) \right ]$

$= (2x + 0) \cdot \left [ - \sin (x^{2} + y^{2}) \right ]$

$= - 2x\sin (x^{2} + y^{2})$

Therefore, $f_{x} (x,y ) = - 2x\sin (x^{2} + y^{2})$ .

Now, find $f_{xy} = \frac{\partial }{\partial y} f_{x}$ as follows:

$f_{x} (x,y ) = - 2x\sin (x^{2} + y^{2})$

$\frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left [ - 2x\sin (x^{2} + y^{2}) \right ]$

$=- 2x \frac{\partial }{\partial y}\left [ \sin (x^{2} + y^{2}) \right ]$

$=- 2x\cdot \frac{\partial }{\partial y}(x^{2} + y^{2}) \cdot \cos (x^{2} + y^{2})$

$=- 2x\cdot (0 + 2y) \cdot \cos (x^{2} + y^{2})$

$=- 2x\cdot ( 2y) \cdot \cos (x^{2} + y^{2})$

$=- 4xy \cos (x^{2} + y^{2})$

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Example Question #1591 : Calculus Ii

Define $f (x,y) = \log xy$ .

Find $f _{xy} (x,y)$ .

Possible Answers:

$f _{xy} (x,y) = \frac{x+y}{x y}$

$f _{xy} (x,y) = 1$

$f _{xy} (x,y) = \frac{1}{x+y}$

$f _{xy} (x,y) = 0$

$f _{xy} (x,y) = \frac{1}{xy}$

Correct answer:

$f _{xy} (x,y) = 0$

Explanation:

First, find $f_{x} = \frac{\partial f}{\partial x}$ as follows:

$f (x,y) = \log xy = \log x + \log y = \frac{ \ln x }{\ln 10}+ \frac{ \ln y }{\ln 10}$

$f (x,y) = \frac{ 1}{\ln 10}\left ( \ln x+ \ln y \right )$

$\frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x}\left [ \frac{ 1}{\ln 10}\left ( \ln x+ \ln y \right ) \right ]$

$= \frac{ 1}{\ln 10} \left (\frac{\partial }{\partial x}\left \ln x+\frac{\partial }{\partial x} \ln y \right )$

$= \frac{ 1}{\ln 10} \left ( \frac{1}{x}+0 \right )$

$= \frac{ 1}{\ln 10} \cdot \frac{1}{x}$

Therefore, $f_{x} (x,y )= \frac{ 1}{\ln 10} \cdot \frac{1}{x}$ .

Now, find $f_{xy} = \frac{\partial }{\partial y} f_{x}$ as follows:

$f_{x} (x,y )= \frac{ 1}{\ln 10} \cdot \frac{1}{x}$

$\frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left ( \frac{ 1}{\ln 10} \cdot \frac{1}{x} \right )$

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Example Question #462 : Derivatives

Let the initial approximation of a solution of the equation

$e^{x} = \sin x$

be $x_{1} = -2$ .

Use one iteration of Newton's method to find an approximation for $x_{2}$ . Give your answer to the nearest thousandth.

Possible Answers:

-4.095

-3.795

-4.195

-3.695

-3.895

Correct answer:

-3.895

Explanation:

Rewrite the equation to be solved for as $e^{x} - \sin x = 0$ .

Let $f(x ) = e^{x} - \sin x$ .

$f'(x ) = \frac{\mathrm{d} }{\mathrm{d} x} \left (e^{x} - \sin x \right )$

$f'(x ) = e^{x} - \cos x \right )$

The problem amounts to finding a zero of f(x ) . By Newton's method, the second approximation can be derived from the first using the equation

$x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})}$ .

Since $x_{1} = -2$ ,

$f(-2 ) = e^{-2} - \sin (-2) \approx 0.1353 - (-0.9093 ) \approx 1.0446$

and

$f'(-2 ) = e^{-2} - \cos (-2) \approx 0.1353 - \left ( -0.4161 \right )\approx 0.5514$

Use these to find the approximation:

$x_{2} =-2 - \frac{f(-2)}{f'(-2)} \approx -2 - \frac{1.0446}{0.5514}\approx -2 -1.8945 \approx -3.895$

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Example Question #467 : Derivative Review

Define f(x) = e^x- 4x .

Give the minimum value of on the set of all real numbers.

Possible Answers:

$4 - 4 \ln 4$

$2- 4 \ln 2$

The function has no miminum value.

e^4 - 16

e^2 - 4

Correct answer:

$4 - 4 \ln 4$

Explanation:

This function is continuous and differentiable everywhere. First we find the value(s) of for which f'(x) =0 .

f(x) = e^x- 4x

$f'(x) =\frac{\mathrm{d} }{\mathrm{d} x}\left ( e^x- 4x \right )$

f'(x) = e^x- 4

f'(x) =0

e^x- 4 =0

e^x= 4

$\ln e^x= \ln 4$

$x= \ln 4$

Therefore, this is the only possible minimum. We determine whether it is a minimum by evaluating $f''(\ln 4)$ :

f'(x) = e^x- 4

$f''(x) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( e^x- 4 \right )$

f''(x) = e^x

$f''(\ln 4) = e^{\ln 4} = 4 > 0$

Since $f''(\ln 4) > 0$ , has its minimum value at $x= \ln 4$ ; it is

$f( \ln 4) = e^{ \ln 4}- 4 \cdot \ln 4 = 4 - 4 \ln 4$ .

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Example Question #1592 : Calculus Ii

Consider the equation

$y =\ln \left ( \sin x + \cos y \right )$ .

Which of the following is equal to $\frac{\mathrm{d} y}{\mathrm{d} x}$ ?

Possible Answers:

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \sin x}{ \sin x + \cos x + \cos y }$

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \sin y + \cos y }$

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \cos x + \cos y }$

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \sin x}{ \sin x + \sin y + \cos y }$

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \sin x + \cos x}{ \sin y + \cos y }$

Correct answer:

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \sin y + \cos y }$

Explanation:

$y =\ln \left ( \sin x + \cos y \right )$

$\frac{\mathrm{d} y }{\mathrm{d} x}= \frac{\mathrm{d} }{\mathrm{d} x}\ln \left ( \sin x + \cos y \right )$

$\frac{\mathrm{d} y }{\mathrm{d} x}= \frac{\frac{\mathrm{d} }{\mathrm{d} x} \left ( \sin x + \cos y \right )}{\sin x + \cos y}$

$\frac{\mathrm{d} y }{\mathrm{d} x} \left ( \sin x + \cos y \right )= \frac{\mathrm{d} }{\mathrm{d} x} \left ( \sin x + \cos y \right )$

$\frac{\mathrm{d} y }{\mathrm{d} x} \sin x + \frac{\mathrm{d} y }{\mathrm{d} x} \cos y \right )= \cos x - \frac{\mathrm{d}y }{\mathrm{d} x} \sin y$

$\frac{\mathrm{d} y }{\mathrm{d} x} \sin x + \frac{\mathrm{d}y }{\mathrm{d} x} \sin y+ \frac{\mathrm{d} y }{\mathrm{d} x} \cos y \right )= \cos x$

$\frac{\mathrm{d} y }{\mathrm{d} x} \left (\sin x + \sin y+ \cos y \right )= \cos x$

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ \cos x}{ \sin x + \sin y+ \cos y }$

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Example Question #1593 : Calculus Ii

Define $f(x) = 4x + 8 \sin x$ .

Give the minimum value of f(x) on the interval $[-\pi, \pi]$ .

Possible Answers:

$-\frac{4}{3} \pi - 4$

$-\frac{8}{3} \pi - 4 \sqrt {3}$

$-4 \pi$

Correct answer:

$-\frac{8}{3} \pi - 4 \sqrt {3}$

Explanation:

$f(x) = 4x + 8 \sin x$

$f'(x) = 4 + 8 \cos x$

We first look for such that f'(x) = 0 :

$4 + 8 \cos x = 0$

$8 \cos x = -4$

$\cos x = -\frac{1}{2}$

The two values on the interval $[-\pi, \pi]$ for which this holds true are $-\frac{2}{3} \pi , \frac{2}{3} \pi$ , so we evaluate for the values $- \pi, -\frac{2}{3} \pi , \frac{2}{3} \pi, \pi$ :

$f(-\pi ) = 4(-\pi ) + 8 \sin (-\pi ) = -4 \pi + 8 \cdot 0 = -4 \pi \approx -12. 5$

$f \left $-\frac{2}{3} \pi\right $ = 4\left $-\frac{2}{3} \pi\right $ + 8 \sin \left $-\frac{2}{3} \pi\right $$

$= -\frac{8}{3} \pi + 8 \cdot \left (- \frac{\sqrt{3}}{2} \right ) = -\frac{8}{3} \pi - 4 \sqrt {3} \approx -15.3$

$f \left $\frac{2}{3} \pi\right $ = 4\left $\frac{2}{3} \pi\right $ + 8 \sin \left $\frac{2}{3} \pi\right $$

$= \frac{8}{3} \pi + 8 \cdot \left ( \frac{\sqrt{3}}{2} \right ) = \frac{8}{3} \pi + 4 \sqrt {3} \approx 15.3$

$f(\pi ) = 4\pi + 8 \sin \pi = 4 \pi + 8 \cdot 0 = 4 \pi \approx 12. 5$

The minimum value is $f \left $-\frac{2}{3} \pi\right $ = -\frac{8}{3} \pi - 4 \sqrt {3}$ .

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Example Question #1594 : Calculus Ii

Consider the equation

$x^2 + y^2 = e^ { xy }$ .

Which of the following is equal to $\frac{\mathrm{d} y}{\mathrm{d} x}$ ?

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x+ y e^ { xy } }{2y- x e^ { xy }}$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x+ 2y }{x e^ { xy }- y e^ { xy }}$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x+ 2y }{y e^ { xy }- x e^ { xy }}$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ 2x- y e^ { xy } }{2y- x e^ { xy }}$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ -2x+ y e^ { xy } }{2y- x e^ { xy }}$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ -2x+ y e^ { xy } }{2y- x e^ { xy }}$

Explanation:

$x^2 + y^2 = e^ { xy }$

$\frac{\mathrm{d} }{\mathrm{d} x}\left ( x^2 + y^2 \right )= \frac{\mathrm{d} }{\mathrm{d} x}e^ { xy }$

$2x + 2y \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} }{\mathrm{d} x} xy \cdot e^ { xy }$

$2x + 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} = \left ( x \cdot \frac{\mathrm{d}y }{\mathrm{d} x} + y \right ) \cdot e^ { xy }$

$2x + 2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} = x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} + y e^ { xy }$

$2y\cdot \frac{\mathrm{d} y}{\mathrm{d} x} - x e^ { xy }\cdot \frac{\mathrm{d}y }{\mathrm{d} x} =-2x+ y e^ { xy }$

$\left (2y- x e^ { xy } \right )\cdot \frac{\mathrm{d} y}{\mathrm{d} x} =-2x+ y e^ { xy }$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{ -2x+ y e^ { xy } }{2y- x e^ { xy }}$

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Example Question #1 : Finding Maximums

Define $f(x) = \frac{x}{2}+ \sin x$ .

Give the maximum value of f(x) on the interval $[0, \pi]$ .

Possible Answers:

$\frac{\pi}{2}$

$\frac{\pi + 2\sqrt{2}}{4}$

$\frac{3\pi + 2\sqrt{2}}{4}$

$\frac{ 2\pi + 3\sqrt{3} }{6}$

$\frac{ \pi + 3\sqrt{3} }{6}$

Correct answer:

$\frac{ 2\pi + 3\sqrt{3} }{6}$

Explanation:

First, we determine if there are any points at which f'(x) = 0 .

$f(x) = \frac{x}{2}+ \sin x = \frac{1}{2}x+ \sin x$

$f'(x) = \frac{1}{2} + \cos x$

f'(x) = 0

$\frac{1}{2} + \cos x = 0$

$\cos x = -\frac{1}{2}$

The only point on the interval on which this is true is $x =\frac{ 2\pi }{3}$ .

We test this point as well as the two endpoints, x = 0 and $x = \pi$ , by evaluating for each of these values.

$f(0) = \frac{0}{2}+ \sin 0 = 0 + 0 = 0$

$f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \left ( \frac{ 2\pi }{3} \right )}{2}+ \sin \frac{ 2\pi }{3} = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} \approx 2$

$f( \pi ) = \frac{ \pi}{2}+ \sin \pi = \frac{ \pi}{2} + 0 = \frac{ \pi}{2} \approx 1.6$

Therefore, assumes its maximum on this interval at the point $x =\frac{ 2\pi }{3}$ , and $f\left ( \frac{ 2\pi }{3} \right ) = \frac{ \pi }{3} + \frac{\sqrt{3}}{2} = \frac{ 2\pi + 3\sqrt{3} }{6}$ .

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Example Question #1591 : Calculus Ii

Consider the equation

$\sec y = x^2 + xy^2$ .

Which of the following is equal to $\frac{\mathrm{d} y}{\mathrm{d} x}$ ?

Possible Answers:

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y - 2xy}$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y + 2xy}$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec ^2 y - 2xy}$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x - y^2}{ \sec y \tan y - 2xy}$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x - y^2}{ \sec ^2 y - 2xy}$

Correct answer:

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y - 2xy}$

Explanation:

$\sec y = x^2 + xy^2$

$\frac{\mathrm{d} }{\mathrm{d} x} \sec y =\frac{\mathrm{d} }{\mathrm{d} x} \left (x^2 + xy^2 \right )$

$\frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y=2x + y^2 + x \cdot 2y \cdot \frac{\mathrm{d}y }{\mathrm{d} x}$

$\frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y=2x + y^2 + 2xy \frac{\mathrm{d}y }{\mathrm{d} x}$

$\frac{\mathrm{d}y }{\mathrm{d} x} \sec y \tan y - \frac{\mathrm{d}y }{\mathrm{d} x} 2xy=2x + y^2$

$\frac{\mathrm{d}y }{\mathrm{d} x} =\frac{2x + y^2}{ \sec y \tan y - 2xy}$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #461 : Derivative Review

Example Question #461 : Derivatives

Example Question #1591 : Calculus Ii

Example Question #462 : Derivatives

Example Question #467 : Derivative Review

Example Question #1592 : Calculus Ii

Example Question #1593 : Calculus Ii

Example Question #1594 : Calculus Ii

Example Question #1 : Finding Maximums

Example Question #1591 : Calculus Ii

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