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Calculus 2 : Derivative at a Point

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Example Questions

Example Question #133 : Derivative Review

What is the slope of a function $f(x)=2x^{2}+13x-7$ at (-1,1) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=2x^{2}+13x-7$ , we can use the Power Rule

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

$f'(x)=(2)2x^{2-1}+(1)13x^{1-1}-(0)7=4x+13$ .

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

f'(-1)=4(-1)+13=-4+13=9 .

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Example Question #11 : Derivative Defined As Limit Of Difference Quotient

Find f'(0) for

$f(x)=x^3\ln(x+1)$

Possible Answers:

$\infty$

Correct answer:

Explanation:

In order to find the derivative, we need to find f'(x) . We can find this by remembering the product rule and knowing 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 this to our problem.

$f(x)=x^3\ln(x+1)$

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

$f'(0)=3(0)^2\ln(0+1)+\frac{0^3}{0+1}=0$

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Example Question #51 : Derivative At A Point

Find the derivative of the following function at the point z=0 :
$f(z)=\frac{z^2\cos(z)}{2}$

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(z)=\frac{4z\cos(z)-2z^2\sin(z)}{4}$

which was found using the following rules:

$\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$ ,

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

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

Finally, plug in the point z=0 into the first derivative function:

$f'(0)=\frac{0}{4}=0$

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Example Question #52 : Derivative At A Point

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 #53 : Derivative At A Point

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

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

Possible Answers:

undefined

$\infty$

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 #54 : Derivative At A Point

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:

$\frac{5}{4}$

$3\tfrac{1}{4}$

$2\tfrac{1}{4}$

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 #51 : Derivative At A Point

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 #140 : Derivative Review

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 #55 : 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 : 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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Calculus 2 : Derivative at a Point

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #133 : Derivative Review

Example Question #11 : Derivative Defined As Limit Of Difference Quotient

Example Question #51 : Derivative At A Point

Example Question #52 : Derivative At A Point

Example Question #53 : Derivative At A Point

Example Question #54 : Derivative At A Point

Example Question #51 : Derivative At A Point

Example Question #140 : Derivative Review

Example Question #55 : Derivative At A Point

Example Question #141 : Derivative Review

All Calculus 2 Resources

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