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Calculus 2 : Derivatives

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Example Questions

Example Question #471 : Derivatives

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:

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

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

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

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

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

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 #472 : Derivatives

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

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

Possible Answers:

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

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

$-4 \pi$

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 #473 : Derivative Review

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:

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

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

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

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

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

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:

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

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

$\frac{\pi}{2}$

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

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

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 #473 : Derivatives

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:

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

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

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

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

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

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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Example Question #1601 : Calculus Ii

Define $f(x) = \frac{1}{6}e^{x} - 6x$ .

Give the minimum value of f(x) on the interval [0,5] .

Possible Answers:

$\frac{1}{6}e - 6$

$6 - 6 \ln 36$

$1 - 6 \ln 6$

$\frac{1}{6}e^{5} - 30$

$\frac{1}{6}$

Correct answer:

$6 - 6 \ln 36$

Explanation:

$f(x) = \frac{1}{6}e^{x} - 6x$

$f'(x) = \frac{1}{6}e^{x} - 6$ .

First, we find out where f'(x) = 0 :

$\frac{1}{6}e^{x} - 6 = 0$

$\frac{1}{6}e^{x} = 6$

$e^{x} = 36$

$x = \ln 36 \approx 3.6$ , which is on the interval.

Now we compare the values of at $x = 0, \ln 36, 5$ :

$f(0) = \frac{1}{6}e^{0} - 6 \cdot 0 = \frac{1}{6} \cdot 1-0 = \frac{1}{6}$

$f(\ln 36) = \frac{1}{6}e^{\ln 36} - 6 \cdot \ln 36 = \frac{1}{6} \cdot 36 - 6 \ln 36 = 6 - 6 \ln 36 \approx -15.5$

$f(0) = \frac{1}{6}e^{5} - 6 \cdot 5 = \frac{1}{6}e^{5} - 30 \approx -5.3$

The answer is $f(\ln 36) = 6 - 6 \ln 36$ .

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Example Question #1602 : Calculus Ii

Consider the equation

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

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

Possible Answers:

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{ e ^{x} -e^y \cos (x+ y) }{ e ^{x} +e^y \cos (x+ y) }$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{ e ^{x} +e^y \cos (x+ y) }{ e ^{x} -e^y \cos (x+ y) }$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{ e ^{x} -e^y \sin (x+ y) }{ e ^{x} +e^y \cos (x+ y) }$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{ e ^{x} +e^y \sin (x+ y) }{ e ^{x} -e^y \sin (x+ y) }$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{ e ^{x} -e^y \sin(x+ y) }{ e ^{x} +e^y \sin (x+ y) }$

Correct answer:

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{ e ^{x} -e^y \cos (x+ y) }{ e ^{x} +e^y \cos (x+ y) }$

Explanation:

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

$\frac{\mathrm{d} }{\mathrm{d} x} \sin (x+ y) = \frac{\mathrm{d} }{\mathrm{d} x}e ^{x-y}$

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

$\left (1 + \frac{\mathrm{d} y }{\mathrm{d} x} \right ) \cdot \cos (x+ y) = \left (1 - \frac{\mathrm{d} y }{\mathrm{d} x} \right ) \cdot e ^{x-y}$

$\cos (x+ y) + \cos (x+ y)\frac{\mathrm{d} y }{\mathrm{d} x} = e ^{x-y} - e ^{x-y} \frac{\mathrm{d} y }{\mathrm{d} x}$

$e ^{x-y} \frac{\mathrm{d} y }{\mathrm{d} x} + \cos (x+ y)\frac{\mathrm{d} y }{\mathrm{d} x} = e ^{x-y} -\cos (x+ y)$

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

$\frac{\mathrm{d} y }{\mathrm{d} x} =\frac{ e ^{x-y} -\cos (x+ y)}{e ^{x-y} + \cos (x+ y) }$

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

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{ e ^{x} -e^y \cos (x+ y) }{ e ^{x} +e^y \cos (x+ y) }$

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Example Question #474 : Derivatives

Define $f (x) = \ln (x + 1) - \ln (x-1)$ .

Give the minimum value of f(x) on the interval [2, 3] .

Possible Answers:

$\ln 4$

$\ln 3$

$\ln 2$

Correct answer:

$\ln 2$

Explanation:

$f (x) = \ln (x + 1) - \ln (x-1)$

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

$=\frac{x - 1}{(x + 1)(x - 1)} - \frac{x + 1}{(x + 1)(x - 1)}$

$=\frac{ - 2}{(x + 1)(x - 1)} = \frac{ - 2}{x^2 - 1}$

Since, on the interval [2, 3] ,

$3 \leq x^2 - 1 \leq 8$ ,

$f' (x) = \frac{ - 2}{x^2 - 1} < 0$ .

is decreasing throughout this interval. Therefore, the minimum of on the interval is

$f (3) = \ln (3 + 1) - \ln (3-1) = \ln 4 - \ln 2 = \ln ( 4 \div 2) = \ln 2$ .

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Example Question #475 : Derivatives

Consider the equation

$x^2 = \cos (xy)$ .

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

Possible Answers:

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{2x + y \sin (xy) }{-x \sin (xy)}$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{2x + y \sin (xy) }{x \cos(xy)}$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{2y + x \cos (xy) }{y \cos(xy)}$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{2x + y \cos (xy) }{x \cos(xy)}$

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{2y + x \cos (xy) }{-y \cos(xy)}$

Correct answer:

$\frac{\mathrm{d} y }{\mathrm{d} x} = \frac{2x + y \sin (xy) }{-x \sin (xy)}$

Explanation:

$x^2 = \cos (xy)$

$\frac{\mathrm{d} }{\mathrm{d} x}x^2 =\frac{\mathrm{d} }{\mathrm{d} x}\cos( xy)$

$2x =\frac{\mathrm{d} }{\mathrm{d} x}(xy)( -\sin (xy))$

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

$2x = - x \frac{\mathrm{d} y }{\mathrm{d} x}\sin (xy) - y \sin (xy)$

$2x + y \sin (xy) = - x \frac{\mathrm{d} y }{\mathrm{d} x}\sin (xy)$

$\frac{2x + y \sin (xy) }{-x \sin (xy)} = \frac{\mathrm{d} y }{\mathrm{d} x}$

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Example Question #1605 : Calculus Ii

Define $f (x) = 4x - \cos x$ .

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

Possible Answers:

Correct answer:

Explanation:

$f (x) = 4x - \cos x$

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

Since $-1 \leq \sin x \leq 1$ ,

$3 \leq 4 + \sin x \leq 5$ ,

and f '(x) is always positive. Therefore, is an always increasing function, and the minimum value of must be $f (0) = 4 \cdot 0 - \cos 0 = 0 - 1 = -1$ .

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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #471 : Derivatives

Example Question #472 : Derivatives

Example Question #473 : Derivative Review

Example Question #1 : Finding Maximums

Example Question #473 : Derivatives

Example Question #1601 : Calculus Ii

Example Question #1602 : Calculus Ii

Example Question #474 : Derivatives

Example Question #475 : Derivatives

Example Question #1605 : Calculus Ii

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