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AP Calculus AB : Derivatives

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Example Questions

Example Question #1 : Slope Of A Curve At A Point

If y = 2x^2+5 , what is the slope of the curve at the point (4,37) ?

Possible Answers:

Correct answer:

Explanation:

To find the slope at a point, we first find the derivative of our function, and then substitute in the -value of our point.

y = 2x^2+5 , so y' = 4x , and plugging in our -value gives 4(4)=16 .

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Example Question #1 : Slope Of A Curve At A Point

Find the rate of change of f(x) at the point (4,12).

f(x)=5x^4-3x^2+15x-17

Possible Answers:

f'(x)=60x^2-6

1271

f'(x)=20x^3-6x+15

34455

Correct answer:

1271

Explanation:

Find the rate of change of f(x) at the point (4,12)

f(x)=5x^4-3x^2+15x-17

We are asked to find a rate of change, so begin by finding the first derivative.

f(x)=5x^4-3x^2+15x-17

f'(x)=20x^3-6x+15

Now, we need the rate of change at (4,12). What matters most is the x value, simply plug it into our derivative and solve for y

f'(4)=20(4)^3-6(4)+15=1280-24+15=1271

So, our answer is 1271

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Example Question #1 : Slope Of A Curve At A Point

Find the slope of the line tangent to f(x) at the point x=0.

f(x)=16x^3+e^x+cos(x)

Possible Answers:

Correct answer:

Explanation:

Find the slope of the line tangent to f(x) at the point x=0

f(x)=16x^3+e^x+cos(x)

To find the slope of a tangent line, we must first find the derivative of our function.

Let's recall a few rules to help us out.

1) The derivative of a monomial can be found by multiplying the coefficient by the exponent, and then decreasing the exponent by 1.

$16x^3\rightarrow (3)16x^2=48x^2$

2) The derivative of e to the x is e to the x

$e^x\rightarrow e^x$

3) The derivative of cosine is negative sine

$cos(x)\rightarrow-sin(x)$

Now, put it all together to get:

f'(x)=48x^2+e^x-sin(x)

Lastly, we need to plug in 0 for x and solve our equation.

$f'(0)=48(0)^2+e^{(0)}-sin(0)=0+1-1=0$

So, our answer is 0

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Example Question #1 : Slope Of A Curve At A Point

$\begin{align*}&\text{For positive values of x: }\\&f(x)=cos(\frac{(\pi x)}{2}) - 5x + sin(4\cdot \pi x) - 5\\&\text{Calculate the slope of the function at }x=6\end{align*}$

Possible Answers:

7.57

-5.33

2.47

Correct answer:

7.57

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }cos(\frac{(\pi x)}{2}) - 5x + sin(4\cdot \pi x) - 5\\&\text{We find the slope function: }4\cdot \pi cos(4\cdot \pi x) -\frac{ (\pi sin(\frac{(\pi x)}{2}))}{2}- 5\\&\text{Plugging in our value of x we then get:}\\&f'(6)=4\cdot \pi cos(4\cdot \pi (6)) -\frac{ (\pi sin(\frac{(\pi (6))}{2}))}{2}- 5\\&f'(6)=7.57\end{align*}$

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Example Question #21 : Derivative At A Point

$\begin{align*}&\text{Calculate the slope of the function:}\\&f(x)=4x + cos(8\cdot \pi x) - sin(\frac{(13\cdot \pi x)}{2}) + 2x^{2} + 3\\&\text{At the point }x=2\end{align*}$

Possible Answers:

32.42

-627.65

Correct answer:

32.42

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }4x + cos(8\cdot \pi x) - sin(\frac{(13\cdot \pi x)}{2}) + 2x^{2} + 3\\&\text{We find the slope function: }4x -\frac{ (13\cdot \pi cos(\frac{(13\cdot \pi x)}{2}))}{2}- 8\cdot \pi sin(8\cdot \pi x) + 4\\&\text{Plugging in our value of x we then get:}\\&f'(2)=4(2) -\frac{ (13\cdot \pi cos(\frac{(13\cdot \pi (2))}{2}))}{2}- 8\cdot \pi sin(8\cdot \pi (2)) + 4\\&f'(2)=32.42\end{align*}$

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Example Question #22 : Derivative At A Point

$\begin{align*}&\text{Calculate the slope of the function:}\\&f(x)=cos(\frac{(13\cdot \pi x)}{2}) + 2sin(4\cdot \pi x) + 2\\&\text{At the point }x=5\end{align*}$

Possible Answers:

0.4

-0.2

4.71

Correct answer:

4.71

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }cos(\frac{(13\cdot \pi x)}{2}) + 2sin(4\cdot \pi x) + 2\\&\text{We find the slope function: }8\cdot \pi cos(4\cdot \pi x) -\frac{ (13\cdot \pi sin(\frac{(13\cdot \pi x)}{2}))}{2}\\&\text{Plugging in our value of x we then get:}\\&f'(5)=8\cdot \pi cos(4\cdot \pi (5)) -\frac{ (13\cdot \pi sin(\frac{(13\cdot \pi (5))}{2}))}{2}\\&f'(5)=4.71\end{align*}$

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Example Question #23 : Derivative At A Point

$\begin{align*}&\text{For positive values of x: }\\&f(x)=2sin(4\cdot \pi x) - cos(4\cdot \pi x) - 5x - 2x^{2} - 3x^{3} + 3\\&\text{Calculate the slope of the function at }x=2\end{align*}$

Possible Answers:

117.91

-23.87

Correct answer:

-23.87

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }2sin(4\cdot \pi x) - cos(4\cdot \pi x) - 5x - 2x^{2} - 3x^{3} + 3\\&\text{We find the slope function: }8\cdot \pi cos(4\cdot \pi x) - 4x + 4\cdot \pi sin(4\cdot \pi x) - 9x^{2} - 5\\&\text{Plugging in our value of x we then get:}\\&f'(2)=8\cdot \pi cos(4\cdot \pi (2)) - 4(2) + 4\cdot \pi sin(4\cdot \pi (2)) - 9(2)^{2} - 5\\&f'(2)=-23.87\end{align*}$

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Example Question #24 : Derivative At A Point

$\begin{align*}&\text{Find the slope of: }\\&f(x)=x + cos(\frac{(3\cdot \pi x)}{2}) + 4cos(\frac{(3\cdot \pi x)}{4}) - sin(7\cdot \pi x) - x^{2} + 3x^{3} - 2\\&\text{At }x=2\end{align*}$

Possible Answers:

56.21

20.43

9.5

Correct answer:

20.43

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }x + cos(\frac{(3\cdot \pi x)}{2}) + 4cos(\frac{(3\cdot \pi x)}{4}) - sin(7\cdot \pi x) - x^{2} + 3x^{3} - 2\\&\text{We find the slope function: }9x^{2} - 7\cdot \pi cos(7\cdot \pi x) -\frac{ (3\cdot \pi sin(\frac{(3\cdot \pi x)}{2}))}{2}- 3\cdot \pi sin(\frac{(3\cdot \pi x)}{4}) - 2x + 1\\&\text{Plugging in our value of x we then get:}\\&f'(2)=9(2)^{2} - 7\cdot \pi cos(7\cdot \pi (2)) -\frac{ (3\cdot \pi sin(\frac{(3\cdot \pi (2))}{2}))}{2}- 3\cdot \pi sin(\frac{(3\cdot \pi (2))}{4}) - 2(2) + 1\\&f'(2)=20.43\end{align*}$

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Example Question #31 : Derivative At A Point

$\begin{align*}&\text{Find the slope of: }\\&f(x)=4x + cos(9\cdot \pi x) - cos(\frac{(11\cdot \pi x)}{2}) + 5x^{2} + x^{3} - 4\\&\text{At }x=2\end{align*}$

Possible Answers:

-1075.99

Correct answer:

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }4x + cos(9\cdot \pi x) - cos(\frac{(11\cdot \pi x)}{2}) + 5x^{2} + x^{3} - 4\\&\text{We find the slope function: }10x - 9\cdot \pi sin(9\cdot \pi x) +\frac{ (11\cdot \pi sin(\frac{(11\cdot \pi x)}{2}))}{2}+ 3x^{2} + 4\\&\text{Plugging in our value of x we then get:}\\&f'(2)=10(2) - 9\cdot \pi sin(9\cdot \pi (2)) +\frac{ (11\cdot \pi sin(\frac{(11\cdot \pi (2))}{2}))}{2}+ 3(2)^{2} + 4\\&f'(2)=36\end{align*}$

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Example Question #32 : Derivative At A Point

$\begin{align*}&\text{Calculate the slope of the function:}\\&f(x)=4x - cos(\frac{(17\cdot \pi x)}{2}) + 3sin(2\cdot \pi x) - sin(6\cdot \pi x) - x^{2} - 3x^{3} - 4\\&\text{At the point }x=9\end{align*}$

Possible Answers:

-248.44

-716.3

-247.89

Correct answer:

-716.3

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }4x - cos(\frac{(17\cdot \pi x)}{2}) + 3sin(2\cdot \pi x) - sin(6\cdot \pi x) - x^{2} - 3x^{3} - 4\\&\text{We find the slope function: }6\cdot \pi cos(2\cdot \pi x) - 2x - 6\cdot \pi cos(6\cdot \pi x) +\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi x)}{2}))}{2}- 9x^{2} + 4\\&\text{Plugging in our value of x we then get:}\\&f'(9)=6\cdot \pi cos(2\cdot \pi (9)) - 2(9) - 6\cdot \pi cos(6\cdot \pi (9)) +\frac{ (17\cdot \pi sin(\frac{(17\cdot \pi (9))}{2}))}{2}- 9(9)^{2} + 4\\&f'(9)=-716.3\end{align*}$

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1 : Slope Of A Curve At A Point

Example Question #1 : Slope Of A Curve At A Point

Example Question #1 : Slope Of A Curve At A Point

Example Question #1 : Slope Of A Curve At A Point

Example Question #21 : Derivative At A Point

Example Question #22 : Derivative At A Point

Example Question #23 : Derivative At A Point

Example Question #24 : Derivative At A Point

Example Question #31 : Derivative At A Point

Example Question #32 : Derivative At A Point

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