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Calculus 2 : Derivative at a Point

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Example Questions

Example Question #31 : Derivative At A Point

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

Possible Answers:

$\frac{1}{3}$

None of the above

Correct answer:

Explanation:

Slope is defined as the first derivative of a function at given point.

We are given the function

$f(x)=x^{2}+7x-3$ and a point (-3,0) , so we need to find the derivative f'(x) and solve for the point's -coordinate.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all nonzero , we can derive

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

Substituting the -coordinate , we have a slope:

f'(-3)=2(-3)+7=-6+7=1 .

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Example Question #111 : Derivatives

What is the slope of a function $f(x)=6x^{2}-12x+9$ at the point (-10,10) ?

Possible Answers:

132

None of the above

Correct answer:

Explanation:

Slope is defined as the first derivative of a function at given point.

We are given the function

$f(x)=6x^{2}-12x+9$ and a point (-10,10) , so we need to find the derivative f'(x) and solve for the point's -coordinate.

Using the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all nonzero , we can derive

$f'(x)=(2)6x^{2-1}-(1)12x^{1-1}+(0)9=12x-12$ .

Substituting the -coordinate , we have a slope:

f'(-10)=12(-10)-12=-120-12=-132 .

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Example Question #112 : Derivatives

What is the slope of $f(x)=4x^{2}-5x+12$ at (1,-1) ?

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the first derivative of a given function.

Since,

$f(x)=4x^{2}-5x+12$ , we can use the Power Rule

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

$f'(x)=(2)4x^{2-1}-(1)5x^{1-1}+(0)12=8x-5$ .

At the point (1,-1) , the -coordinate is .

Thus, the slope is

f'(1)=8(1)-5=8-5=3 .

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Example Question #111 : Derivatives

What is the slope of $f(x)=x^{2}+2x+11$ at (3,0) ?

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the first derivative of a given function.

Since,

$f(x)=x^{2}+2x+11$ , we can use the Power Rule

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

$f'(x)=(2)x^{2-1}+(1)2x^{1-1}+(0)11=2x+2$ .

At the point (3,0) , the -coordinate is .

Thus, the slope is

f'(3)=2(3)+2=6+2=8 .

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Example Question #114 : Derivatives

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

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the first derivative of a given function.

Since,

$f(x)=7x^{2}-6$ , we can use the Power Rule

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

$f'(x)=(2)7x^{1-1}-(0)6=14x$ .

At the point (5,3) , the -coordinate is .

Thus, the slope is f'(5)=14(5)=70 .

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Example Question #1241 : Calculus Ii

What is the slope of a function f(x)=5x^2-x+7 at the point (4,5) ?

Possible Answers:

None of the above

Correct answer:

Explanation:

Slope is defined as the first derivative of a given function.

Since f(x)=5x^2-x+7 , we can use the Power Rule

$\frac{d}{dx}x^n=nx^{n-1}$ for all $n\neq 0$ to determine that

$f'(x)=(2)5x^{2-1}-(1)x^{1-1}+(0)7=10x-1.$

Since we're given a point (4,5) , we can use the -coordinate to solve for the slope at that point.

Thus,

f'(4)=10(4)-1=40-1=39 .

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Example Question #13 : Derivative Defined As Limit Of Difference Quotient

What is the slope of a function f(x)=-2x^2+5x-3 at the point (0,1) ?

Possible Answers:

Correct answer:

Explanation:

Slope is defined as the first derivative of a given function.

Since f(x)=-2x^2+5x-3 , we can use the Power Rule

$\frac{d}{dx}x^n=nx^{n-1}$ for all $n\neq 0$ to determine that

$f'(x)=(2)(-2)x^{2-1}+(1)5x^{1-1}-(0)3=-4x+5$ .

Since we're given a point (0,1) , we can use the x-coordinate to solve for the slope at that point.

Thus,

f'(0)=-4(0)+5=5.

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Example Question #116 : Derivatives

What is the slope of a function f(x)=-x^2+8x-4 at the point (-1,0) ?

Possible Answers:

$-\frac{1}{10}$

$\frac{1}{10}$

None of the above

Correct answer:

Explanation:

Slope is defined as the first derivative of a given function.

Since f(x)=-x^2+8x-4 , we can use the Power Rule

$\frac{d}{dx}x^n=nx^{n-1}$ for all $n\neq 0$ to determine that

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

Since we're given a point (-1,0) , we can use the x-coordinate to solve for the slope at that point.

Thus,

f'(-1)=-2(-1)+8=2+8=10 .

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Example Question #1241 : Calculus Ii

What is the slope of the tangent line to the function

$h(x)=e^{x}ln(x)$

when x = 1?

Possible Answers:

$\frac{1}{e}$

Correct answer:

Explanation:

The slope of the tangent line to a function at a point is the value of the derivative at that point. To calculate the derivative in this problem, the product rule is necessary. Recall that the product rule states that:

$\frac{d}{dx}(m(x)n(x))=m'(x)n(x)+m(x)n'(x)$ .

In this example, $m(x)=e^x \ and\ n(x)=ln(x).$

Therefore,

$m'(x)=e^x \ and\ n'(x)=\frac{1}{x}$ , and

$\frac{d}{dx}(e^xln(x))=e^xln(x)+e^x\cdot\frac{1}{x}$

At x = 1, this dervative has the value

$e^1ln(1)+e^1\cdot\frac{1}{1}=0+e=e$ .

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Example Question #32 : Derivative At A Point

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

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

Possible Answers:

Correct answer:

Explanation:

The derivative of the function is

$f'(x)=-2\cos(x)\sin(x)+3$

and was found using the following rules:

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

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

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

where $f(x)=\cos^2(x), g(x)=x$ in the chain rule.

Plug in 0 in the derivative function to get f'(x)=3

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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 #31 : Derivative At A Point

Example Question #111 : Derivatives

Example Question #112 : Derivatives

Example Question #111 : Derivatives

Example Question #114 : Derivatives

Example Question #1241 : Calculus Ii

Example Question #13 : Derivative Defined As Limit Of Difference Quotient

Example Question #116 : Derivatives

Example Question #1241 : Calculus Ii

Example Question #32 : Derivative At A Point

All Calculus 2 Resources

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