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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 #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:

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

$\frac{1}{6}$

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

$6 - 6 \ln 36$

$1 - 6 \ln 6$

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 #1602 : 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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Example Question #1 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Find the derivative of: y=cos^-^1(x)+cos(x)

Possible Answers:

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

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

$\frac{cos(x)\sqrt{1-x^2}-sin(x)}{\sqrt{1-x^2}}$

$\frac{1+cos(x)}{\sqrt{1+x^2}}$

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

Correct answer:

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

Explanation:

The derivative of inverse cosine is:

$\frac{d}{dx} cos^-^1(x) = -\frac{1}{\sqrt{1-x^2}}$

The derivative of cosine is:

$\frac{d}{dx} cos(x)=-sin(x)$

Combine the two terms into one term.

$-\frac{1}{\sqrt{1-x^2}}-sin(x)$

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

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Example Question #3 : Other Derivative Review

The position function of a car is modeled as s(t)=4x^2 + 8x +1 .

Find the speed of the car at t=2 . HINT: The derivative of the position function is the speed function.

Possible Answers:

Correct answer:

Explanation:

As hinted in the problem statement, we need to find the derivative.

Using the simple power rule for derivatives $f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ we find the derivative to be,

s'(t)=v(t)=8x+8 .

Now, plug in t=2 to solve.

v(2)=16+8=24

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Example Question #4 : Other Derivative Review

$f(x)=e^{3x}sin\left(\frac{x}{3}\right)$

Calculate f'(x) .

Possible Answers:

$3e^{3x}cos\left(\frac{x}{3}\right)+\frac{1}{3}e^{3x}sin\left(\frac{x}{3}\right)$

$3e^{3x}sin{\frac{x}{3}}+\frac{1}{3}e^{3x}cos\left(\frac{x}{3}\right)$

$e^{3x}sin\left(\frac{x}{3}\right)+e^{3x}cos\left(\frac{x}{3}\right)$

$3e^{3x}sin\left(\frac{x}{3}\right)+e^{3x}cos\left(\frac{x}{3}\right)$

$e^{3x}cos\left(\frac{x}{3}\right)$

Correct answer:

$3e^{3x}sin{\frac{x}{3}}+\frac{1}{3}e^{3x}cos\left(\frac{x}{3}\right)$

Explanation:

To calculate this derivate we need to use the product rule, as we have two functions multiplied by each other.

The product rule is defined as

$\frac{d}{dx}f(x)\cdot g(x)=f'(x)\cdot g(x) + f(x)\cdot g'(x)$

Let's make $e^{3x}$ be our f(x) and $sin\left(\frac{x}{3}\right)$ be our g(x) .

We get

$3e^{3x}sin{\frac{x}{3}}+\frac{1}{3}e^{3x}cos\left(\frac{x}{3}\right)$ .

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Example Question #3 : Other Derivative Review

$y=2x^{e^{x}}$ . Find ${\frac{\mathrm{d} y}{\mathrm{d} x}}$

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}\cdot e^{x}(\frac{1}{x}+\ln\left | x \right |)$

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

$\frac{\mathrm{d} y}{\mathrm{d} x}=2e^{x}x^{e^{x}-1}$

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}e^{x}\cdot \ln\left | x \right |$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}\cdot e^{x}(\frac{1}{x}+\ln\left | x \right |)$

Explanation:

$x^{e^{x}}$ is a variable raised to a variable power. There is no no way to find ${\frac{\mathrm{d} y}{\mathrm{d} x}}$ explicityly as it follows no derivative formula. So we must start by rearranging the original equation, $y=2x^{e^{x}}$ , to a point that we can implicitly differentiate.

First isolate the $x^{e^{x}}$ , by dividing both sides by .

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

Now take the natural log of both sides of the equation.

$\ln\left |\frac{y}{2} \right |=\ln\left |x^{e^{x}} \right |$ .

This enables the use of log properties, specifically the property $\ln (m^{n})=n\cdot \ln(m)$ . Applying this property to the right side of the equation gives the following.

$\ln\left |\frac{y}{2} \right |=e^{x}\cdot \ln\left |x\right |$

Now that there isn't a variable raised to a variable power anymore, so we can differentiate implicitly without issue.

The left side follows the pattern, $\frac{\mathrm{d} }{\mathrm{d} u}(\ln\left | u \right |)=\frac{du}{u}$ .We use the product rule on the right hand side of the equation.

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

$\frac{\frac{1}{2}\frac{\mathrm{d} y}{\mathrm{d} x}}{\frac{y}{2}}=(e^{x})(\frac{1}{x})+(e^{x})(\ln\left | x \right |)$

Now solve for ${\frac{\mathrm{d} y}{\mathrm{d} x}}$ . First, simplify the left side of the equation, and pull the greatest common factor out of the right side.

$\frac{\frac{\mathrm{d} y}{\mathrm{d} x}}y=e^{x}(\frac{1}{x}+\ln\left | x \right |)$

Now multiply both sides by to cancel it off the right side.

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

Next, using the original equation, $y=2x^{e^{x}}$ , replace with $2x^{e^{x}}$ . this puts everything back in terms of x.

$\frac{\mathrm{d} y}{\mathrm{d} x}=2x^{e^{x}}\cdot e^{x}(\frac{1}{x}+\ln\left | x \right |)$

This is the final answer.

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Example Question #6 : Other Derivative Review

Differentiate: y=cot(3x)

Possible Answers:

$-\frac{1}{3}csc^2(3x)$

-cos^2(3x)

-3cos^2(3x)

-csc^2(3x)

-3csc^2(3x)

Correct answer:

-3csc^2(3x)

Explanation:

In order to differentiate cotangent, write the rule for the derivative.

$\frac{d}{dx} cot(x) = -csc^2(x)$

We will also need to use chain rule and multiply the derivative of the inner function.

$\frac{d}{dx} cot(3x) = -csc^2(3x) \cdot 3$

The answer is: -3csc^2(3x)

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #1601 : Calculus Ii

Example Question #1602 : Calculus Ii

Example Question #474 : Derivatives

Example Question #475 : Derivatives

Example Question #1602 : Calculus Ii

Example Question #1 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #3 : Other Derivative Review

Example Question #4 : Other Derivative Review

Example Question #3 : Other Derivative Review

Example Question #6 : Other Derivative Review

All Calculus 2 Resources

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