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AP Calculus AB : Derivatives

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Example Questions

Example Question #1203 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=2x + 6x^{2} + 10\text{ as h approaches 0 for }x=-10\end{align*}$

Possible Answers:

580

$-\frac{59}{3}$

590

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=2x + 6x^{2} + 10\text{ and we can find that }\\&f(x+h)=2\cdot h + 2x + 6\cdot (h + x)^{2} + 10\\&f'(x)=\frac{2\cdot h + 2x + 6\cdot (h + x)^{2} + 10-(2x + 6x^{2} + 10)}{h}\\&f'(x)=\frac{2\cdot h + 12\cdot hx + 6\cdot h^{2}}{h}\\&f'(x)=6\cdot h + 12x + 2\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-10:\\&-118\end{align*}$

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Example Question #1204 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=- 10x - 6\text{ as h approaches 0 for }x=-1\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=- 10x - 6\text{ and we can find that }\\&f(x+h)=- 10\cdot h - 10x - 6\\&f'(x)=\frac{- 10\cdot h - 10x - 6-(- 10x - 6)}{h}\\&f'(x)=\frac{-10\cdot h}{h}\\&f'(x)=-10\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-1:\\&-10\end{align*}$

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Example Question #1205 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=5x^{2} - 4\text{ as h approaches 0 for }x=-9\end{align*}$

Possible Answers:

401

$\text{Nonexistent.}$

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=5x^{2} - 4\text{ and we can find that }\\&f(x+h)=5\cdot (h + x)^{2} - 4\\&f'(x)=\frac{5\cdot (h + x)^{2} - 4-(5x^{2} - 4)}{h}\\&f'(x)=\frac{10\cdot hx + 5\cdot h^{2}}{h}\\&f'(x)=5\cdot h + 10x\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-9:\\&-90\end{align*}$

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Example Question #1206 : Ap Calculus Ab

Find the derivative of the function using the limit of the difference quotient:

f(x)=12x+3

Possible Answers:

The limit does not exist

None of the other answers

Correct answer:

Explanation:

The derivative of a function, f(x) , as defined by the limit of the difference quotient is

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Taking the limit of our function - and remembering the limit at each step! - we get

$\lim_{h\rightarrow 0}\frac{12(x+h)+3-(12x+3)}{h}=\lim_{h\rightarrow 0}\frac{12x+12h+3-12x-3}{h}=\lim_{h\rightarrow 0}\frac{12h}{h}=12$

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Example Question #1207 : Ap Calculus Ab

Find the derivative of the function using the limit of the difference quotient:

f(x)=3x^2+x

Possible Answers:

6x+3h+1

6x+1

Correct answer:

6x+1

Explanation:

The derivative of a function, f(x) , as defined by the limit of the difference quotient is

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

Taking the limit of our function (and remembering to write the limit at each step) we get

$\lim_{h\rightarrow 0}\frac{3(x+h)^2+(x+h)-(3x^2+x)}{h}=\lim_{h\rightarrow 0}\frac{3(x^2+2xh+h^2)+x+h-3x^2-x}{h}=\lim_{h\rightarrow 0}\frac{3x^2+6xh+3h^2+x+h-3x^2-x}{h}=\lim_{h\rightarrow 0}\frac{h(6x+3h+1)}{h}=6x+1$

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Example Question #51 : Concept Of The Derivative

Find the derivative of the function using the limit of the difference quotient:

f(x)=5x^2+1

Possible Answers:

10x+1

10x

15xh

10x+5h

Correct answer:

10x

Explanation:

The derivative of a function f(x) is defined by the limit of the difference quotient, as follows:

$\lim_{h\rightarrow 0}\frac{ f(x+h)-f(x)}{h}$

Using this limit for our function, and remembering to write the limit at every step, we get

$\lim_{h\rightarrow 0} \frac{5(x+h^2)+1-(5x^2+1)}{h}=\lim_{h\rightarrow 0} \frac{5(x^2+2xh+h^2)+1-5x^2-1}{h}=\lim_{h\rightarrow 0}\frac{10xh+5h^2}{h}=\lim_{h\rightarrow 0} (10x+5h) = 10x$

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Example Question #2121 : High School Math

Let f(x)=sin(x)-cos(x) .

Find the second derivative of f(x) .

Possible Answers:

f''(x)=sin^2(x)-cos^2(x)

f''(x)=sin(x)-cos(x)

f''(x)=cos(x)-sin(x)

f''(x)=sin^2(x)+cos^2(x)

f''(x)=sin(x)+cos(x)

Correct answer:

f''(x)=cos(x)-sin(x)

Explanation:

The second derivative is just the derivative of the first derivative. So first we find the first derivative of f(x) . Remember the derivative of $\sin x$ is $\cos x$ , and the derivative for $\cos x$ is $-\sin x$ .

f'(x)= cos(x)+sin(x)

Then to get the second derivative, we just derive this function again. So

f''(x)=-sin(x)+cos(x)=cos(x)-sin(x)

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Example Question #51 : Calculus I — Derivatives

Define $f (x) = 6x^{3} - 12x^{2} + 4x -8$ .

What is f ' ' (x) ?

Possible Answers:

f ' '(x) = 18

f ' '(x) = 36x

f ' '(x) = 6x - 1 2

f ''(x) = 18x - 24

f ' '(x) = 36x - 24

Correct answer:

f ' '(x) = 36x - 24

Explanation:

Take the derivative of , then take the derivative of .

$f (x) = 6x^{3} - 12x^{2} + 4x -8$

$f '(x) = 3 \cdot 6x^{3-1} - 2 \cdot 12x^{2-1} + 4 -0$

$f '(x) = 18x^{2} - 24x+ 4$

$f ' '(x) = 2 \cdot 18x^{2-1} - 24+ 0$

f ' '(x) = 36x - 24

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Example Question #12 : Finding Second Derivative Of A Function

Define $g(x) = \cos \left (x^{2} \right )$ .

What is g''(x) ?

Possible Answers:

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

$g''(x) = - 2 \cos(x^2)-4x^2 \sin(x^2) \right ]$

$g''(x) =-4x^2 \cos(x^2)$

$g''(x) = -4x^2 \sin(x^2)$

$g''(x) =4x^2 \cos(x^2)$

Correct answer:

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

Explanation:

Take the derivative of , then take the derivative of .

$g(x) = \cos \left (x^{2} \right )$

$g'(x) = 2x \cdot [-\sin (x^2)]$

$g'(x) = - 2x \sin (x^2)$

$g''(x) = - 2\left [ 1 \cdot \sin(x^2)+ x \cdot 2x \cos(x^2) \right ]$

$g''(x) = - 2\left [ \sin(x^2)+ 2x^2 \cos(x^2) \right ]$

$g''(x) = - 2 \sin(x^2)-4x^2 \cos(x^2) \right ]$

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Example Question #11 : Finding Second Derivative Of A Function

Define $f(x) = \frac{1}{e^{4x}}$ .

What is f ''(x) ?

Possible Answers:

$f(x) = \frac{1}{16e^{4x}}$

$f''(x) = \frac{1 }{e^{4x-2}}$

$f''(x) = \frac{16 }{e^{4x}}$

$f''(x) =- \frac{16 }{e^{4x}}$

$f''(x) = \frac{1 }{e^{4x+2}}$

Correct answer:

$f''(x) = \frac{16 }{e^{4x}}$

Explanation:

Rewrite:

$f(x) = \frac{1}{e^{4x}}$

$f(x) = e^{-4x}$

Take the derivative of , then take the derivative of .

$f'(x) = -4 e^{-4x}$

$f''(x) = -4 \cdot \left (-4 \right )e^{-4x} = 16e^{-4x} = \frac{16 }{e^{4x}}$

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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 #1203 : Ap Calculus Ab

Example Question #1204 : Ap Calculus Ab

Example Question #1205 : Ap Calculus Ab

Example Question #1206 : Ap Calculus Ab

Example Question #1207 : Ap Calculus Ab

Example Question #51 : Concept Of The Derivative

Example Question #2121 : High School Math

Example Question #51 : Calculus I — Derivatives

Example Question #12 : Finding Second Derivative Of A Function

Example Question #11 : Finding Second Derivative Of A Function

All AP Calculus AB Resources

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