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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 #132 : Derivative Review

Find the derivative of the following function at $\theta=2\pi$ :

$f(\theta )=\sin^3(\theta)+e^{\theta}$

Possible Answers:

$e^{2\pi}$

$3+2\pi e^{2\pi}$

Correct answer:

$e^{2\pi}$

Explanation:

The derivative of the function is

$f'(\theta)=3\sin^2(\theta)\cos(\theta)+ e^{\theta}$

and was found using the following rules:

$\frac{d}{dx}(e^{u})=e^{u}\frac{du}{dx}$ ,

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

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

$\frac{d}{dx}(x^n)=nx^{^{n-1}}$

To finish the problem, plug in $\theta=2 \pi$ into the derivative function:

$f'(2 \pi)= e^{2\pi}$ .

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Example Question #133 : Derivative Review

Find the derivative of the following function at $y=2 \pi$

$f(y)=\sec^2(y)+\tan(y)$ .

Possible Answers:

$\infty$

undefined

Correct answer:

Explanation:

The derivative of the function is

$f'(y)=2\sec^2(y)\tan(y)+\sec^2(y)$

and was found using the following rules:

$\frac{d}{dx}\tan(x)=\sec^2(x)$ , $\frac{d}{dx} \sec(x)=\sec(x)\tan(x)$ , $\frac{d}{dx} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{d}{dx}(x^n)=nx^{n-1}$

Then, plug in the point given into the first derivative function:

$f'(2 \pi) = 2(1)(0)+1=1$

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Example Question #1261 : Calculus Ii

Find the second derivative of the following function at x=0 :

$f(x)=\cos(x)+e^\frac{x}{2}+x^3+x^2$

Possible Answers:

$2\tfrac{1}{4}$

$\frac{5}{4}$

$3\tfrac{1}{4}$

Correct answer:

$\frac{5}{4}$

Explanation:

We must find the first derivative of the function first:

$f'(x)=-\sin(x)+\frac{e^{\frac{x}{2}}}{2}+3x^2+2x$

The derivative was found using the following rules:

$\frac{d}{dx}\cos(x)=-\sin(x)$ , $\frac{d}{dx}(e^u)=e^u\frac{du}{dx}$ , $\frac{d}{dx}(x^n)=nx^{n-1}$

Find the second derivative of the function by taking the derivative of the above function:

$f''(x)=-\cos(x)+\frac{e^{\frac{x}{2}}}{4}+6x+2$

An additional rule was used:

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

Now, plug in x=0 into the above function:

$f''(0)=-1+\frac{1}{4}+2=\frac{5}{4}$

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Example Question #1261 : Calculus Ii

Find the second derivative of the following function at x=0 :

$e^{2x}+2x\cos(3x)$

Possible Answers:

Correct answer:

Explanation:

To find the second derivative of the function, we first must find the first derivative of the function:

$f'(x)=2e^{2x}+2\cos(3x)-6x\sin(3x)$

The derivative was found using the following rules:

$\frac{d}{dx}\cos(x)=-\sin(x)$ , $\frac{d}{dx}e^u=e^u\frac{du}{dx}$ , $\frac{d}{dx}(x^n)=nx^{n-1}$ , $\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$

The second derivative is simply the derivative of the first derivative function, and is equal to:

$f''(x)=4e^{2x}-12\sin(3x)-18x\cos(3x)$

One more rule used in combination with some of the ones above is:

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

To finish the problem, plug in x=0 into the above function to get an answer of .

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Example Question #52 : Derivative At A Point

What is the slope of $f(x)=9x^{2}-12x+5$ at (4,-4) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=9x^{2}-12x+5$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to determine that

$f'(x)=(2)9x^{2-1}-(1)12x+(0)5=18x-12$ .

We also have a point (4,-4) with a -coordinate , so the slope

f'(4)=18(4)-12=72-12=60 .

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Example Question #141 : Derivative Review

What is the slope of $f(x)=10x^{2}+x-4$ at (2,7) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=10x^{2}+x-4$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to determine that

$f'(x)=(2)10x^{2-1}+(1)x^{1-1}-(0)4=20x+1$ .

We also have a point (2,7) with a -coordinate , so the slope

f'(2)=20(2)+1=40+1=41 .

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Example Question #141 : Derivative Review

What is the slope of $f(x)=x^{2}-3x+7$ at (1,-7) ?

Possible Answers:

Correct answer:

Explanation:

We define slope as the first derivative of a given function.

Since we have

$f(x)=x^{2}-3x+7$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to determine that

$f'(x)=(2)x^{2-1}-(1)3x^{1-1}+(0)7=2x-3$ .

We also have a point (1,-7) with a -coordinate , so the slope

f'(1)=2(1)-3=2-3=-1 .

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Example Question #61 : Derivative At A Point

Find f'(1) of $f(x)=x^2+\ln(x)$ .

Possible Answers:

Correct answer:

Explanation:

In order to take the derivative, we need to use the power rule and the definition of the derivative of natural log.

Remember that the derivative of natural log is:

$\\ f(x)=\ln(g(x)) \\ \\ f'(x)=\frac{g'(x)}{g(x)}$

Remember that the power rule is:

$\\ f(x)=x^a \\ f'(x)=ax^{a-1}$

Now lets apply these rules to this problem.

$f(x)=x^2+\ln(x)$

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

Now we simply plug in 1.

$f'(1)=2(1)+\frac{1}{1}=3$

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Example Question #62 : Derivative At A Point

Find f'(0) for

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

Possible Answers:

$\frac{1}{2}$

Correct answer:

Explanation:

In order to find f'(0) , we must first find f'(x) .

In order to find f'(x) , we need to remember the product rule and the derivative of natural log.

Product Rule:

[f(x)g(x)]'=f'(x)g(x)+f(x)g'(x)

Derivative of natural log:

$\\ f(x)=\ln(g(x)) \\ \\ f'(x)=\frac{g'(x)}{g(x)}$

Now lets apply these rules to our problem.

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

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

$f'(0)=\log(1)+\frac{0}{1}=0+0=0$

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Example Question #141 : Derivatives

Find the derivative of the following function at x=0 :

$f(x)=e^{2x}\cos^2(x)+2x^2+3$

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is:

$f'(x=2e^{2x}\cos^2(x)-2e^{2x}\cos(x)\sin(x)+4x$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \sin(x)=-\cos(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$

To finish the problem, plug in zero into the function above. We get an answer of 2.

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #132 : Derivative Review

Example Question #133 : Derivative Review

Example Question #1261 : Calculus Ii

Example Question #1261 : Calculus Ii

Example Question #52 : Derivative At A Point

Example Question #141 : Derivative Review

Example Question #141 : Derivative Review

Example Question #61 : Derivative At A Point

Example Question #62 : Derivative At A Point

Example Question #141 : Derivatives

All Calculus 2 Resources

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