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AP Calculus BC : Derivative at a Point

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Example Question #1 : Derivative At A Point

Find the value of the following derivative at the point $\left(2,\frac{1}{2}\right)$ :

$f(x)=\frac{1}{\sqrt{x+2}}$

Possible Answers:

$\frac{1}{4}$

$-\frac{1}{16}$

$\frac{1}{16}$

${}-\frac{1}{4}$

$\frac{1}{\sqrt{\frac{5}{2}}}$

Correct answer:

$-\frac{1}{16}$

Explanation:

To solve this problem, first we need to take the derivative of the function. It will be easier to rewrite the equation as $f(x)=(x+2)^{-\frac{1}{2}}$ from here we can take the derivative and simplify to get

$\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{1}{2}-1}\cdot \frac{\mathrm{d} }{\mathrm{d} x}(x+2)\\ f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}\cdot 1\\f'(x)=-\frac{1}{2}(x+2)^{-\frac{3}{2}}$

From here we need to evaluate at the given point $\left(2,\frac{1}{2}\right)$ . In this case, only the x value is important, so we evaluate our derivative at x=2 to get $f'(2)=-\frac{1}{2}(2+2)^{-\frac{3}{2}}=-\frac{1}{2}\cdot \frac{1}{8}=-\frac{1}{16}$ .

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Example Question #3 : Derivative At A Point

Evaluate the value of the derivative of the given function at the point (-1,4) :

f(x)=3x^4+2x^2-1

Possible Answers:

Correct answer:

Explanation:

To solve this problem, first we need to take the derivative of the function.

$\\f'(x)=4\cdot 3x^{4-1}+2\cdot 2x^{2-1}\\f'(x)=12x^3+4x$

From here we need to evaluate at the given point (-1,4) . In this case, only the x value is important, so we evaluate our derivative at x=1 to get

$\\f'(-1)=12(-1)^3+4(-1)\\ f'(-1)=12(-1)+4(-1)\\ f'(-1)=-16$ .

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Example Question #7 : Derivative Defined As Limit Of Difference Quotient

Given $f(x)=5x^{2}+6x-12$ , find the value of f'(x) at the point (-1,-1) .

Possible Answers:

Correct answer:

Explanation:

Given the function $f(x)=5x^{2}+6x-12$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)5x^{2-1}+(1)6x^{1-1}-(0)12=10x+6$ .

Plugging in the -value of the point (-1,-1) into f'(x) , we get

f'(-1)=10(-1)+6=-10+6=-4 .

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Example Question #14 : Derivative At A Point

Given $f(x)=10x^{2}-9x+1$ , find the value of f'(x) at the point (-2,7) .

Possible Answers:

Correct answer:

Explanation:

Given the function $f(x)=10x^{2}-9x+1$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all $n\neq0$ to find its derivative:

$f'(x)=(2)10x^{2-1}-(1)9x^{1-1}+(0)=20x-9$ .

Plugging in the -value of the point (-2,7) into f'(x) , we get

f'(-2)=20(-2)-9=-40-9=-49 .

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Example Question #15 : Derivative At A Point

Find the derivative of y=(2x+4)(3x+2) at point (1,30) .

Possible Answers:

Correct answer:

Explanation:

Use either the FOIL method to simplify before taking the derivative or use the product rule to find the derivative of the function.

The product rule will be used for simplicity.

y'=(2x+4)(3)+(3x+2)(2) = 6x+12 + 6x+4= 12x+16

Substitute x=1 .

y'(1)= 12(1)+16=28

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Example Question #103 : Derivatives

Find the derivative of the following function at x=2 :

$h(x)=2x\sin(x)$

Possible Answers:

$2\sin(2)+4\cos(2)$

$2\sin(2)-4\cos(2)$

$2\cos(2)$

$2\sin(2)$

Correct answer:

$2\sin(2)+4\cos(2)$

Explanation:

The derivative of the function is given by the product rule:

h(x)=f(x)g(x) , h'(x)=f'(x)g(x)+f(x)g'(x)

Simply find the derivative of each function:

$\frac{d}{dx}(2x)=2$

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

The derivatives were found using the following rules:

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

Simply evaluate each derivative and the original functions at the point given, using the above product rule.

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Example Question #37 : Derivative At A Point

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 #11 : Derivative At A Point

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 #52 : Derivative At A Point

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 #57 : 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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AP Calculus BC : Derivative at a Point

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #1 : Derivative At A Point

Example Question #3 : Derivative At A Point

Example Question #7 : Derivative Defined As Limit Of Difference Quotient

Example Question #14 : Derivative At A Point

Example Question #15 : Derivative At A Point

Example Question #103 : Derivatives

Example Question #37 : Derivative At A Point

Example Question #11 : Derivative At A Point

Example Question #52 : Derivative At A Point

Example Question #57 : Derivative At A Point

All AP Calculus BC Resources

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