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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

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

Find the slope of $f(t)=\sqrt{5t+6}$ at t=0.

Possible Answers:

$\frac{5}{2\sqrt{6}}$

$\sqrt{6}$

$\frac{-5}{2\sqrt{6}}$

$\frac{7}{2\sqrt{6}}$

Correct answer:

$\frac{5}{2\sqrt{6}}$

Explanation:

First, use the chain rule to find f'(t).

You should get $f'(t)=\frac{5}{2}(5t+6)^{-\frac{1}{2}}$ .

Next, plug in t=0 to get f'(0)= $\frac{5}{2\sqrt{6}}$ .

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

Find the slope of the curve 3x^2 , at x=4 .

Possible Answers:

m=12

m=48

m=16

m=24

m=-24

Correct answer:

m=24

Explanation:

Find the derivative of the curve using the power rule.

In mathematical terms, the power rule states,

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

Therefore the derivative is,

$\frac{d}{dx}3x^2\rightarrow 2\cdot 3x^{2-1}=6x$

Next, plug in the x-value to find the slope of the curve at x=4 , which gives you a final answer of m=24.

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Example Question #2 : Slope Of A Curve At A Point

Find the slope of the curve, (x+6)^2+3 , at x=0 .

Possible Answers:

m=12

Correct answer:

m=12

Explanation:

Expand the binomial and combine like terms to get,

x^2+12x+39.

Next, take the derivative of the polynomial using the power rule, which is in mathematical terms,

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

Therefore,

$\frac{d}{dx}x^2+12x+39\rightarrow 2x^{2-1}+1\cdot 12x^{1-1}=2x+12$

Lastly, plug in for , to get the slope at the point.

This gives you your final answer of m=12.

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

Find the slope at the point (1,2) of the function

y = x^3 + 2x -3

Possible Answers:

= 3

= 5

= 4

= 6

= -5

Correct answer:

= 5

Explanation:

Derivatives are slope finders. Therefore to find the slope at the given point, we need to find the derivative of the function using power rule. Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. $n(x^{n-1})$

So, we get y' =3x^2 + 2

From there we plug in our x value.

3(1)^2 +2 = 5

slope=5

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

Find the slope of the function $y = xe^{2x}$ at the point $(\ln2,4\ln2)$

Possible Answers:

$8\ln2+4$

$\frac{\ln2}{4}+8$

None of the other answers

$4\ln2$

Correct answer:

$8\ln2+4$

Explanation:

To find the slope at a point of our function, we need to find its derivative first.

Using the product rule (and the chain rule within this product rule application), we have

$y' = (x)(2e^{2x})+(e^{2x})(1) = (2x+1)e^{2x}$ .

Plugging the -value of our point into this equation, we get our desired slope of

$(2\ln2+1)e^{2\ln2} = (2\ln2+1)e^{\ln4}=(2\ln2+1)4 = 8\ln2 +4$ .

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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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AP Calculus AB : AP Calculus AB

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 #2 : 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 #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

All AP Calculus AB Resources

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