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Calculus 2 : Derivative at a Point

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Example Question #41 : Derivative At A Point

What is the slope of $f(x)=x^{2}-6x+15$ at (2,-4) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=x^{2}-6x+15$ , we can use the Power Rule

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

$f'(x)=(2)x^{2-1}-(1)6x^{1-1}+(0)15=2x-6$ .

We also have a point (2,-4) with a -coordinate , so the slope

f'(2)=2(2)-6=4-6=-2 .

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Example Question #42 : Derivative At A Point

What is the slope of $f(x)=4x^{2}+x-1$ at (1,2) ?

Possible Answers:

None of the above

Correct answer:

Explanation:

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

Since we have

$f(x)=4x^{2}+x-1$ , we can use the Power Rule

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

$f'(x)=(2)4x^{2-1}+(1)x^{1-1}-(0)1=8x+1$ .

We also have a point (1,2) with a -coordinate , so the slope

f'(1)=8(1)+1=8+1=9 .

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Example Question #43 : Derivative At A Point

What is the slope of $f(x)=-x^{2}-5x-7$ at (4,-4) ?

Possible Answers:

None of the above

Correct answer:

Explanation:

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

Since we have

$f(x)=-x^{2}-5x-7$ , we can use the Power Rule

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

$f'(x)=(2)(-1)x^{2-1}-(1)5x^{1-1}-(0)7=-2x-5$ .

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

f'(4)=-2(4)-5=-8-5=-13 .

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Example Question #3 : Derivative Rules For Sums, Products, And Quotients

Find the slope of the tangent line to the function $h(x)=\frac{2ln(x)}{x^2}$ at x=1 .

Possible Answers:

ln(2)-1

Correct answer:

Explanation:

The slope of the tangent line to a function at a point is the value of the derivative of the function at that point. In this problem, h(x) is a quotient of two functions, $2ln(x) \ and\ x^2$ , so the quotient rule is needed.

In general, the quotient rule is

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

To apply the quotient rule in this example, you must also know that $\frac{d}{dx}ln(x)=\frac{1}{x}$ and that $\frac{d}{dx}x^n=nx^{n-1}$ .

Therefore, the derivative is

$h'(x)=\frac{x^2(\frac{2}{x})-2ln(x)(2x)}{(x^2)^2}=\frac{2x-4xln(x)}{x^4}$

The last step is to substitute for in the derivative, which will tell us the slope of the tangent line to h(x) at x = 1 .

$h'(1)=\frac{2(1)-4(1)ln(1)}{1^4}=\frac{2-0}{1}=2$

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Example Question #41 : Derivative At A Point

What is the slope of $f(x)=x^{2}-3x+7$ at the point (9,0) ?

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 (9,0) with a -coordinate , so the slope

f'(9)=2(9)-3=18-3=15 .

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Example Question #45 : Derivative At A Point

What is the slope of $f(x)=\frac{x^{2}}{2}+12x-5$ at the point (-3,3) ?

Possible Answers:

Correct answer:

Explanation:

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

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

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

$f'(x)=2\left(\frac{x^{2-1}}{2}\right)+(1)12x^{1-1}-(0)5=x+12$ .

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

f'(-3)=(-3)+12=9 .

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Example Question #46 : Derivative At A Point

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

Possible Answers:

Correct answer:

Explanation:

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

Since we have $f(x)=3x^{2}-2x+1$ , we can use the Power Rule

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

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

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

f'(1)=6(1)-2=6-2=4 .

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Example Question #47 : Derivative At A Point

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

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

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

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

and was found using the following rules:

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

where f(x)=2x , $g(x)=\cos^2(x)$ , and $\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

Simply plug in x=0 into the first derivative function and solve:

f'(0)=2+0=2

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Example Question #41 : Derivative At A Point

What is the slope of a function $f(x)=3x^{2}+7x-5$ at (2,6) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=3x^{2}+7x-5$ , we can use the Power Rule

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

$f'(x)=(2)3x^{2-1}+(1)7x^{1-1}-(0)5=6x+7$ .

We also have a point (2,6) with a -coordinate , so the slope

f'(2)=6(2)+7=12+7=19 .

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Example Question #49 : Derivative At A Point

What is the slope of a function $f(x)=5x^{2}-9x+14$ at (-4,2) ?

Possible Answers:

Correct answer:

Explanation:

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

Since we have

$f(x)=5x^{2}-9x+14$ , we can use the Power Rule

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

$f'(x)=(2)5x^{2-1}-(1)9x^{1-1}+(0)14=10x-9$ .

We also have a point (-4,2) with a -coordinate , so the slope

f'(-4)=10(-4)-9=-40-9=-49 .

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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 #41 : Derivative At A Point

Example Question #42 : Derivative At A Point

Example Question #43 : Derivative At A Point

Example Question #3 : Derivative Rules For Sums, Products, And Quotients

Example Question #41 : Derivative At A Point

Example Question #45 : Derivative At A Point

Example Question #46 : Derivative At A Point

Example Question #47 : Derivative At A Point

Example Question #41 : Derivative At A Point

Example Question #49 : Derivative At A Point

All Calculus 2 Resources

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