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Calculus 2 : Derivatives

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Example Questions

Example Question #52 : Derivative At A Point

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 : Derivatives

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 #143 : Derivative Review

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 #144 : Derivative Review

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 #145 : Derivative Review

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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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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Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #52 : Derivative At A Point

Example Question #141 : Derivatives

Example Question #143 : Derivative Review

Example Question #144 : Derivative Review

Example Question #145 : Derivative Review

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

All Calculus 2 Resources

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