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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #461 : Derivative Review

Consider the equation

2x^2 + y^2 = 8xy

where is a function of .

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

Possible Answers:

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

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

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{2x- 4 y}{4x- y}$

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{4x- 2 y}{4x- y}$

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{4x- y}{2x- 4 y}$

Correct answer:

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{2x- 4 y}{4x- y}$

Explanation:

2x^2 + y^2 = 8xy

$\frac{\mathrm{d} (2x^2 + y^2) }{\mathrm{d} x}=\frac{\mathrm{d} \left (8xy) \right }{\mathrm{d} x}$

$2\; \frac{\mathrm{d} }{\mathrm{d} x} x^2+ \frac{\mathrm{d} }{\mathrm{d} x}y^2=8\: \frac{\mathrm{d} \left \right }{\mathrm{d} x}xy$

$2 \cdot 2x+ 2y \cdot \frac{\mathrm{d}y }{\mathrm{d} x} =8 \left ( y + x \cdot \frac{\mathrm{d} y}{\mathrm{d} x}\right )$

$4x+ 2y \cdot \frac{\mathrm{d}y }{\mathrm{d} x} = 8 y + 8x \cdot \frac{\mathrm{d} y}{\mathrm{d} x}$

$4x- 8 y = 8x \cdot \frac{\mathrm{d} y}{\mathrm{d} x} - 2y \cdot \frac{\mathrm{d}y }{\mathrm{d} x}$

$\left ( 8x- 2y \right )\cdot \frac{\mathrm{d}y }{\mathrm{d} x} = 4x- 8 y$

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{4x- 8 y}{8x- 2y}$

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{2x- 4 y}{4x- y}$

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Example Question #461 : Derivative Review

Use implicit differentiation to find $\frac{dy}{dx}$ :

$x^4\sin(x+y)=2$

Possible Answers:

$\frac{dy}{dx} = -\left(4\tan(x+y) +1 \right )$

$\frac{dy}{dx} = \frac{4\tan(x+y)}{x}$

$\frac{dy}{dx} = -\left(\frac{4\sin(x+y)-2x}{x\cos x}\right)$

$\frac{dy}{dx} = 2-4x^3\sin (x+y)$

$\frac{dy}{dx}=-\left(\frac{4\tan(x+y)+x}{x} \right )$

Correct answer:

$\frac{dy}{dx}=-\left(\frac{4\tan(x+y)+x}{x} \right )$

Explanation:

To differentiate the left-hand side of the equation, we must use the product rule: f'(x)g(x) +f(x)g'(x) .

Let f(x)=x^4 so that f'(x)=4x^3 , and $g(x)=\sin(x+y)$ so that ${g'(x)=\cos(x+y) \cdot \left(1+\frac{dy}{dx} \right )}$ . After differentiation, we end up with ${4x^3\sin(x+y) +x^4\cos(x+y)\left(1+\frac{dy}{dx} \right )=0}$ .

Using algebra to isolate the $\frac{dy}{dx}$ term, we find that ${\frac{dy}{dx}= -\frac{4x^3 \sin(x+y)}{x^4\cos(x+y)}-1} =-\left(\frac{4\tan(x+y) +x}{x} \right )$ .

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Example Question #462 : Derivative Review

Find $\frac{dy}{dx}$ :

$y=x^{\sin x}$

Possible Answers:

$\frac{dy}{dx}= \ln\left( x^{\sin x}\right ) \cos x$

$\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x} \right )$

$\frac{dy}{dx} = -\cos x \left( x^{\sin x}\right)$

$\frac{dy}{dx} = \cos x \ln x +\frac{\sin x }{x} \right$

$\frac{dy}{dx} = \sin x\left(x^{\cos x} \right )$

Correct answer:

$\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x} \right )$

Explanation:

Since $\tiny{x}$ is a part of both the base and the exponent, we need to use logarithmic differentiation; that is, take the log of both sides of the equation: $\ln y = \ln (x^{\sin x}) \rightarrow \ln y =\sin x \ln x.$

Differentiating the latter equation, we obtain $\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x +\sin x \cdot \left(\frac{1}{x}\right)$ .

Thus, $\frac{dy}{dx} = x^{\sin x}\left(\cos x \ln x + \frac{\sin x}{x} \right )$ .

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Example Question #462 : Derivative Review

Define $f (x,y) = 3x^{2} \ln y$ .

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

Possible Answers:

$f _{xy} (x,y) = 6xy$

$f _{xy} (x,y) = \frac{6}{xy}$

$f _{xy} (x,y) = \frac{6x }{y}$

$f _{xy} (x,y) = \frac{6y }{x}$

$f _{xy} (x,y) = \frac{6}{xy^{2}}$

Correct answer:

$f _{xy} (x,y) = \frac{6x }{y}$

Explanation:

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

$f (x,y) = 3x^{2} \ln y$

$\frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} 3x^{2} \ln y$

$= 3 \ln y \cdot \frac{\partial }{\partial x} x^{2}$

$= 3 \ln y \cdot 2x$

$= 6x \ln y$

Therefore, $f_{x} (x,y ) = 6x \ln y$ .

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

$f_{x} (x,y ) = 6x \ln y$

$\frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y} 6x \ln y$

$= 6x \cdot \frac{\partial }{\partial y} \ln y$

$= 6x \cdot \frac{1}{y} = \frac{6x }{y}$

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Example Question #463 : Derivative Review

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

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

Possible Answers:

$f_{x} (x,y ) = \sec ^{2} xy + 2 x y \sec ^{2} xy \tan xy$

$f_{x} (x,y ) = \sec xy \tanxy + 2 x y \sec ^{2}xy$

$f_{x} (x,y ) = \sec ^{2} xy + 2 x y \sec ^{3} xy$

$f_{x} (x,y ) = 2 x y \sec ^{2} xy \tan xy$

$f_{x} (x,y ) = \sec xy \tan xy + 2 x y \sec ^{2} xy \tan xy$

Correct answer:

$f_{x} (x,y ) = \sec ^{2} xy + 2 x y \sec ^{2} xy \tan xy$

Explanation:

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

$f (x,y) = \tan xy$

$\frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x} \tan xy$

$= \frac{\partial }{\partial x} xy \cdot \sec ^{2} xy$

$= y \sec ^{2} xy$

Therefore, $f_{x} (x,y ) = y \sec ^{2} xy$ .

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

$f_{x} (x,y ) = y \sec ^{2} xy$

$f_{x} (x,y ) = \frac{\partial y}{\partial y} \cdot \sec ^{2} xy + y \cdot \frac{\partial }{\partial y} \sec ^{2}xy$

$f_{x} (x,y ) = 1 \cdot \sec ^{2} xy + y \cdot 2 \cdot \frac{\partial }{\partial y}\sec xy \cdot \sec xy$

$f_{x} (x,y ) = \sec ^{2} xy + 2 y \cdot \frac{\partial }{\partial y} xy \cdot \sec xy \tan xy \cdot \sec xy$

$f_{x} (x,y ) = \sec ^{2} xy + 2 y \cdot x \cdot \sec xy \tan xy \cdot \sec xy$

$f_{x} (x,y ) = \sec ^{2} xy + 2 x y \sec ^{2} xy \tan xy$

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Example Question #464 : Derivative Review

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

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

Possible Answers:

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

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

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

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

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

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 #464 : Derivative Review

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

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

Possible Answers:

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

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

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

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

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

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:

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

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

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

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

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

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 #1592 : Calculus Ii

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:

-3.895

-4.195

-3.695

-3.795

-4.095

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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Calculus 2 : Derivative Review

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #461 : Derivative Review

Example Question #461 : Derivative Review

Example Question #462 : Derivative Review

Example Question #462 : Derivative Review

Example Question #463 : Derivative Review

Example Question #464 : Derivative Review

Example Question #464 : Derivative Review

Example Question #1591 : Calculus Ii

Example Question #1592 : Calculus Ii

Example Question #467 : Derivative Review

All Calculus 2 Resources

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