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AP Calculus AB : Derivative at a point

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Example Question #129 : Derivatives

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

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 #130 : Derivatives

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

Possible Answers:

-247.89

-716.3

-248.44

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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Example Question #11 : Slope Of A Curve At A Point

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

Possible Answers:

-0.03

-13.97

0.82

$\infty$

Correct answer:

-13.97

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }ln(3x) + sin(\frac{(9\cdot \pi x)}{2}) + 2\\&\text{We find the slope function: }\frac{(9\cdot \pi cos(\frac{(9\cdot \pi x)}{2}))}{2}+\frac{ 1}{x}\\&\text{Plugging in our value of x we then get:}\\&f'(6)=\frac{(9\cdot \pi cos(\frac{(9\cdot \pi (6))}{2}))}{2}+\frac{ 1}{(6)}\\&f'(6)=-13.97\end{align*}$

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Example Question #12 : Slope Of A Curve At A Point

$\begin{align*}&\text{Calculate the slope of the function:}\\&f(x)=2ln(3x) - 4x + sin(5\cdot \pi x) - sin(\frac{(9\cdot \pi x)}{2}) + 3x^{2} + 6x^{3} + 2\\&\text{At the point }x=1\end{align*}$

Possible Answers:

8.2

6.29

239.86

$\infty$

Correct answer:

6.29

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }2ln(3x) - 4x + sin(5\cdot \pi x) - sin(\frac{(9\cdot \pi x)}{2}) + 3x^{2} + 6x^{3} + 2\\&\text{We find the slope function: }6x + 5\cdot \pi cos(5\cdot \pi x) -\frac{ (9\cdot \pi cos(\frac{(9\cdot \pi x)}{2}))}{2}+\frac{ 2}{x}+ 18x^{2} - 4\\&\text{Plugging in our value of x we then get:}\\&f'(1)=6(1) + 5\cdot \pi cos(5\cdot \pi (1)) -\frac{ (9\cdot \pi cos(\frac{(9\cdot \pi (1))}{2}))}{2}+\frac{ 2}{(1)}+ 18(1)^{2} - 4\\&f'(1)=6.29\end{align*}$

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Example Question #13 : Slope Of A Curve At A Point

$\begin{align*}&\text{For positive values of x: }\\&f(x)=3x^{3} - cos(7\cdot \pi x) - 6\\&\text{Calculate the slope of the function at }x=2\end{align*}$

Possible Answers:

8.5

519.61

Correct answer:

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }3x^{3} - cos(7\cdot \pi x) - 6\\&\text{We find the slope function: }7\cdot \pi sin(7\cdot \pi x) + 9x^{2}\\&\text{Plugging in our value of x we then get:}\\&f'(2)=7\cdot \pi sin(7\cdot \pi (2)) + 9(2)^{2}\\&f'(2)=36\end{align*}$

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Example Question #14 : Slope Of A Curve At A Point

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

Possible Answers:

-3.11

15.85

-2.89

2.47

Correct answer:

15.85

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }3sin(2\cdot \pi x) - 3x - sin(\frac{(\pi x)}{2}) + 2\\&\text{We find the slope function: }6\cdot \pi cos(2\cdot \pi x) -\frac{ (\pi cos(\frac{(\pi x)}{2}))}{2}- 3\\&\text{Plugging in our value of x we then get:}\\&f'(9)=6\cdot \pi cos(2\cdot \pi (9)) -\frac{ (\pi cos(\frac{(\pi (9))}{2}))}{2}- 3\\&f'(9)=15.85\end{align*}$

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Example Question #31 : Derivative At A Point

$\begin{align*}&\text{Find the slope of: }\\&f(x)=2e^{(\frac{x}{3})} - 6x + cos(\pi x) - sin(\pi x) + 2\\&\text{At }x=10\end{align*}$

Possible Answers:

-3.64

9.55

-0.09

-0.59

Correct answer:

9.55

Explanation:

$\begin{align*}&\text{Recall the definition of slope as rise over run. To put it in}\\&\text{other words, it is a change in height in relation to a change}\\&\text{horizontal position: it is a rate. And since it is a rate, it}\\&\text{can be found with a derivative. In this case, the derivative}\\&\text{of our function (at a height at a given point) with respect}\\&\text{to x:}\\&Slope=\frac{df(x)}{dx}\\&\text{Taking the derivative of our function, }2e^{(\frac{x}{3})} - 6x + cos(\pi x) - sin(\pi x) + 2\\&\text{We find the slope function: }\frac{(2e^{(\frac{x}{3})})}{3}- \pi cos(\pi x) - \pi sin(\pi x) - 6\\&\text{Plugging in our value of x we then get:}\\&f'(10)=\frac{(2e^{(\frac{(10)}{3})})}{3}- \pi cos(\pi (10)) - \pi sin(\pi (10)) - 6\\&f'(10)=9.55\end{align*}$

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Example Question #32 : Derivative At A Point

Find the slope of the line tangent to when t=3

v(t)=13t^2-12t+e^t-6

Possible Answers:

96.09

86.09

54.07

95.01

Correct answer:

86.09

Explanation:

Find the slope of the line tangent to when t=3

v(t)=13t^2-12t+e^t-6

Now, let's first look at this conceptually. We need to find the slope of a tangent line at a particular value for t. This sounds like a job for a derivative.

We will first find v'(t), then we will plug in three to find v'(3), which will yield our answer.

First, recall:

1) $\frac{d}{dx}e^x=e^x$

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

These two rules are all that we need to solve this problem:

v(t)=13t^2-12t+e^t-6

v'(t)=(2)13t-12+e^t=26t-12+e^t

So, we have:

v'(t)=26t-12+e^t

Let's plug in 3...

v'(3)=26(3)-12+e^3=86.09

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Example Question #33 : Derivative At A Point

Find the slope of the line tangent to the curve of d(t) when t is equal to 0.

d(t)=3t^2-6sin(t)+12cos(t)-11t

Possible Answers:

Correct answer:

Explanation:

Find the slope of the line tangent to the curve of d(t) when t is equal to 0.

d(t)=3t^2-6sin(t)+12cos(t)-11t

So, we are asked to find the slope of a tangent line at a particular point. To do so, we need to find the derivative of our function, and then evaluate it at the given value of t.

Let's begin by finding d'(t).

To do so, we need to use the power rule, as well as the rules for sine and cosine.

1) Power Rule

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

2) Sine Rule

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

3) Cosine Rule

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

Now, we can use all of these to find the derivative of d(t)

d(t)=3t^2-6sin(t)+12cos(t)-11t

Becomes:

d'(t)=6t-6cos(t)-12sin(t)-11

Now, we are almost there, but we need to evaluate this derivative when t=0.

d'(0)=6(0)-6cos(0)-12sin(0)-11

This simplifies quite nicely to:

d'(0)=0-6+0-11=-17

So, the slope of our tangent line when t=0 is negative 17.

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Example Question #131 : Derivatives

If $f(x)=|x-1|+3$ x=1 , what is the value of $f'(1)$ ?

Possible Answers:

$\infty$

Does not exist.

Correct answer:

Does not exist.

Explanation:

The function is not differentiable at x=1 . So the derivative at x=1 does not exist.

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AP Calculus AB : Derivative at a point

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #129 : Derivatives

Example Question #130 : Derivatives

Example Question #11 : Slope Of A Curve At A Point

Example Question #12 : Slope Of A Curve At A Point

Example Question #13 : Slope Of A Curve At A Point

Example Question #14 : Slope Of A Curve At A Point

Example Question #31 : Derivative At A Point

Example Question #32 : Derivative At A Point

Example Question #33 : Derivative At A Point

Example Question #131 : Derivatives

All AP Calculus AB Resources

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