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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

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9 Diagnostic Tests 308 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1581 : Calculus Ii

Find the second derivative of f(x)=2x^4-3x^3+5x^2-1 .

Possible Answers:

$f{}''(x)=4x^2-18x+10$

$f{}''(x)=24x^2-18x$

$f{}''(x)=24x^2-18x+10$

$f{}''(x)=24x^2-18x-1$

$f{}''(x)=24x^2+10$

Correct answer:

$f{}''(x)=24x^2-18x+10$

Explanation:

Before finding the second derivative, you must find the first derivative. Remember to multiply the exponent by the coefficient and then subtract one from the exponent. Evaluate each term separately and then put all together at the end.

The first derivative is: $f{}'(x)=8x^3=9x^2+10x$ .

Now, take the derivative of the 1st derivative to find the second derivative:

$f{}''(x)=24x^2-18x+10$ .

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Example Question #1582 : Calculus Ii

Find the first derivative of $f(x)=\frac{6x^3-8x^2}{x^2}$ .

Possible Answers:

f{}'(x)=6

$f{}'(x)=\frac{6}{x}$

$f{}'(x)=6x^2$

f{}'(x)=6x

f{}'(x)=6x+1

Correct answer:

f{}'(x)=6

Explanation:

First, chop up the fraction into 2 separate terms and simplify:

$\frac{6x^3-8x^2}{x^2}=\frac{6x^3}{x^2}-\frac{8x^2}{x^2}=6x-8$

Now, take the derivative of that expression. Remember to multiply the exponent by the coefficient and then decrease the exponent by 1:

f{}'(x)=6

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Example Question #1583 : Calculus Ii

$\int \frac{5}{x}-1dx$

Possible Answers:

$5ln\left | x\right |-x+C$

$5ln\left | x\right |-2x+C$

$5ln\left | x\right |-x$

$5ln\left | x\right |-5x+C$

$ln\left | x\right |-x+C$

Correct answer:

$5ln\left | x\right |-x+C$

Explanation:

Integrate each term separately. When there is just an x on the denominator, the integral of that is $ln\left | x\right |$ .

Therefore, the first term integrated is:

$5ln\left | x\right |$

The second term integrated is:

Put those together to get:

$5ln\left | x\right |-x$

Because it's an indefinite integral, remember to add C at the end:

$5ln\left | x\right |-x+C$ .

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Example Question #3 : Velocity, Speed, Acceleration

Given the velocity function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

where is real number such that $d\in [0,1]$ , find the acceleration function

a(x) .

Possible Answers:

$a(x)=(x+1)^{d}\sin (x+1)^{10}$

$a(x)=(2x+2)^{d}(x+1)^{2d}\sin (x+1)^{10}$

$a(x)=(x^2+2x+1)^{d}\sin (x^2+2x+1)^{10}$

$a(x)=(x^2+2x+1)^{2d}\sin (x^2+2x+1)^{10}$

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Correct answer:

$a(x)=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

Explanation:

We can find the acceleration function a(x) from the velocity function by taking the derivative:

a(x)=v'(x)

We can view the function

$v(x)=\int_{0}^{x^2+2x+1} z^d\sin(z^5) dz$

as the composition of the following functions

$g(x)=\int_{0}^{x} z^d\sin(z^5) dz$

h(x)=x^2+2x+1=(x+1)^2

so that v(x)=g(h(x)) . This means we use the chain rule

[g(h(x))]'=g'(h(x))h'(x)

to find the derivative. We have $g'(x)=x^d\sin x^5$ and h'(x)=2x+2 , so we have

$a(x)=v'(x)=[(x+1)^2]^d\sin [(x+1)^2]^5\cdot (2x+2)$

$=(2x+2)(x+1)^{2d}\sin (x+1)^{10}$

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Example Question #451 : Derivatives

Define $f (x,y) = \frac{1}{x^{2}y^{2}}$ .

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

Possible Answers:

$f _{xy} (x,y) = - \frac{9}{x^{3}y^{3} }$

$f _{xy} (x,y) = \frac{1}{x^{3}y^{3} }$

$f _{xy} (x,y) = \frac{4}{x^{3}y^{3} }$

$f _{xy} (x,y) = \frac{9}{x^{3}y^{3} }$

$f _{xy} (x,y) = - \frac{4}{x^{3}y^{3} }$

Correct answer:

$f _{xy} (x,y) = \frac{4}{x^{3}y^{3} }$

Explanation:

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

$f (x,y) = \frac{1}{x^{2}y^{2}}$

$\frac{\partial }{\partial x}f (x,y) = \frac{\partial }{\partial x}\left ( \frac{1}{x^{2}y^{2}} \right )$

$=\frac{1}{ y^{2}} \cdot \frac{\partial }{\partial x}\left ( \frac{1}{x^{2} } \right )$

$=\frac{1}{ y^{2}} \cdot \frac{\partial }{\partial x}x^{-2}$

$=\frac{1}{ y^{2}} \cdot (-2) x^{-3} = -\frac{2}{x^{3}y^{2}}$

Therefore, $f_{x} (x,y ) = -\frac{2}{x^{3}y^{2}}$

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

$f_{x} (x,y ) = -\frac{2}{x^{3}y^{2}}$

$\frac{\partial }{\partial y} f_{x} (x,y ) = \frac{\partial }{\partial y}\left ( -\frac{2}{x^{3}y^{2}} \right )$

$= -\frac{2}{x^{3} } \cdot \frac{\partial }{\partial y}\left ( \frac{1}{ y^{2}} \right )$

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

$= -\frac{2}{x^{3} } \cdot (-2) y^{-3}$

$= -\frac{2}{x^{3} } \cdot \left ( -\frac{2}{y^{3} } \right )$

$= \frac{4}{x^{3}y^{3} }$

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Example Question #461 : Derivatives

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:

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

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

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

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

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

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 #1 : Other Derivative Review

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

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

Possible Answers:

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

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

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

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

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

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 #2 : Other Derivative Review

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

$y=x^{\sin x}$

Possible Answers:

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

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

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

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

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

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 : Derivatives

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

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

Possible Answers:

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

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

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

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

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

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 : Derivatives

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

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

Possible Answers:

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

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

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

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

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

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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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1581 : Calculus Ii

Example Question #1582 : Calculus Ii

Example Question #1583 : Calculus Ii

Example Question #3 : Velocity, Speed, Acceleration

Example Question #451 : Derivatives

Example Question #461 : Derivatives

Example Question #1 : Other Derivative Review

Example Question #2 : Other Derivative Review

Example Question #462 : Derivatives

Example Question #463 : Derivatives

All Calculus 2 Resources

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