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Calculus 2 : Calculus II

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Example Questions

Example Question #142 : Derivatives

Find the derivative of the function at x=0 :

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

Possible Answers:

undefined

Correct answer:

Explanation:

The first derivative of the function is

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

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ ,

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

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

$\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

To finish the problem, plug in x=0 into the first derivative equation:

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

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Example Question #147 : Derivative Review

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

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

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(x)=\sec(x)\tan(x)+2e^{2x}$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x}f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x}\sec(x)=\sec(x)\tan(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$

To finish, plug in the given point into the first derivative function:

f'(0)=0+2=2

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Example Question #148 : Derivative Review

What is the derivative of the function $f(x) = x \ln x$ at the point x = e ?

Possible Answers:

e + 1

Correct answer:

Explanation:

Using the product rule which states,

$f(x)g(x)\rightarrow f(x)g'(x)+g(x)f'(x)$

and the definition for the derivative of natural log,

$ln(u)\rightarrow \frac{1}{u}\cdot \frac{du}{dx}$

for the function gives you

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

Substituting x = e and you get $\ln e + 1$ .

Recall that the natural log of is just , and that leads us to our answer: .

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Example Question #141 : Derivative Review

Find the derivative of $(\ln x)^{x}$ at the point x = e .

Possible Answers:

Undefined

Correct answer:

Explanation:

Since we can't difrerentiate that problem as it stands, we can rewrite it using properties of logs to make it easier to derive. $(\ln x)^x$ can be rewritten as $x \ln(\ln x)$ using the properties of logs. Now this is much easier to derive. Use the product rule and treat as one part and $\ln(\ln x)$ as another part. Taking the derivative of the first part and leaving the second part gives us just $\ln(\ln x)$ . Leaving the first part and taking the derivative of the second part is a little trickier. To do that, we first must know the derivative of $\ln(\ln x)$ which we can figure out via the chain rule.

Using the chain rule, we get

$\frac{1}{\ln x} \cdot \frac{1}{x}$ as the derivative.

The we originally had cancels out the $\frac{1}{x}$ (using the second part of the product rule) so our final derivative becomes

$\ln(\ln x) + \frac{1}{\ln x}$ .

Substituting x = e in gives us

$\ln(\ln e) + \frac{1}{\ln e}$ .

Recall that the natural log of e is just 1, and the natural log of 1 is 0, which leads us $0 + \frac{1}{1}$ which gives us our final answer of .

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Example Question #61 : Derivative At A Point

What is the slope of $f(x)=-7x^{2}+x-12$ at (3,4) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=-7x^{2}+x-12$ , we can use the Power Rule

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

$f'(x)=(2)(-7)x^{2-1}+(1)x^{1-1}-(0)12=-14x+1$

We also have a point (3,4) with a -coordinate , so the slope

f'(3)=-14(3)+1=-42+1=-41 .

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Example Question #151 : Derivative Review

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

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=x^{2}+4x+5$ , we can use the Power Rule

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

$f'(x)=(2)x^{2-1}+(1)4x^{1-1}+(0)5=2x+4$ .

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

f'(2)=2(2)+4=4+4=8 .

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Example Question #151 : Derivative Review

What is the slope of $f(x)=3x^{2}-5x+6$ at (5,-3) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=3x^{2}-5x+6$ , we can use the Power Rule

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

$f'(x)=(2)3x^{2-1}-(1)5x^{1-1}+(0)6=6x-5$ .

We also have a point (5,-3) with a -coordinate , so the slope

f'(5)=6(5)-5=30-5=25 .

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Example Question #152 : Derivative Review

Find f'(0) for

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

Possible Answers:

$\frac{\pi}{4}$

$\frac{\pi}{2}$

$\pi$

Correct answer:

Explanation:

In order to take the derivative, we need to remember how to take derivative of natural log and the product rule.

Derivative of natural log:

$f(t)=\ln(g(t))$

$f'(t)=\frac{g'(t)}{g(t)}$

Product Rule:

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

Now lets apply both of these rule to our problem.

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

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

We plug in 0 to get.

$f'(0)=\cos(0)\ln(1)+\frac{\sin(0)}{1}$

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

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Example Question #151 : Derivative Review

Find f'(1) for

$f(x)=x^2\cos(x)$

Possible Answers:

$2\cos(1)$

$2\tan(1)$

$2\cos(1)+\sin(1)$

$2\cos(1)-\sin(1)$

The derivative doesn't exist.

Correct answer:

$2\cos(1)-\sin(1)$

Explanation:

In order to find f'(1) , we first find f'(x) .

To find the derivative, we need to remember the product rule.

Product Rule:

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

Lets apply this rule to our problem.

$f(x)=x^2\cos(x)$

$f'(x)=2x\cos(x)-x^2\sin(x)$

We plug in 1 to get

$f'(1)=2(1)\cos(1)-(1)^2\sin(1)$

$=2\cos(1)-\sin(1)$

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Example Question #153 : Derivative Review

Find f'(0) for

$f(x)=2^{2x+2}$

Possible Answers:

$8\ln(2)$

$\log(2)$

Correct answer:

$8\ln(2)$

Explanation:

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

We need to remember the derivative of expotential functions.

Derivative of expotential functions:

$f(x)=a^{g(x)}$

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

Now lets apply this to our problem.

$f(x)=2^{2x+2}$

$f'(x)=2*2^{2x+2}*\ln(2)$

$f'(0)=2*2^{2(0)+2}*\ln(2)=8\ln(2)$

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Calculus 2 : Calculus II

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #142 : Derivatives

Example Question #147 : Derivative Review

Example Question #148 : Derivative Review

Example Question #141 : Derivative Review

Example Question #61 : Derivative At A Point

Example Question #151 : Derivative Review

Example Question #151 : Derivative Review

Example Question #152 : Derivative Review

Example Question #151 : Derivative Review

Example Question #153 : Derivative Review

All Calculus 2 Resources

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