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AP Calculus AB : Concept of the Derivative

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Example Questions

Example Question #21 : Concept Of The Derivative

Find the derivative of the following function:

f(x)=5x+2

Possible Answers:

Correct answer:

Explanation:

The derivative of a function f(x) can be defined according to the following:

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

Using this for our function - and not forgetting to write the limit for every step! - we get

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

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Example Question #13 : Derivative Defined As The Limit Of The Difference Quotient

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

f(x)=2x^2+3

Possible Answers:

The limit does not exist

4x+2h

Correct answer:

Explanation:

The derivative of a function as defined by the limit of the difference quotient is

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

Evaluating the limit using our function - and always writing the limit symbol! - we get

$\lim_{h\rightarrow 0}\frac{2(x+h)^2+3-(2x^2+3)}{h}= \lim_{h\rightarrow 0}\frac{2(x^2+2xh+h^2)+3-2x^2-3}{h}=\lim_{h\rightarrow 0}\frac{4xh+2h^2}{h}=\lim_{h\rightarrow0}4x+2h=4x$

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Example Question #14 : Derivative Defined As The Limit Of The Difference Quotient

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

f(x)=x^2+2x+1

Possible Answers:

2x+h+2

2x+2

2x^2+1

Correct answer:

2x+2

Explanation:

To find the derivative using the limit of the difference quotient, we use the formula:

$f'=\lim_{h\rightarrow 0}\left [ \frac{f(x+h)-f(x)}{h}\right ]$

$f'=\lim_{h\rightarrow 0}\left [ \frac{(x^2+2xh+h^2)+2(x+h)+1-(x^2+2x+1) }{h}\right ]$

$f'=\lim_{h\rightarrow 0}\frac{2xh+h^2+2h}{h}=2x+2$

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Example Question #11 : Derivative Defined As The Limit Of The Difference Quotient

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

Possible Answers:

Correct answer:

Explanation:

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

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Example Question #1181 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=-8x\text{ as h approaches 0.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=-8x\text{ and we can find that }\\&f(x+h)=- 8\cdot h - 8x\\&f'(x)=\frac{- 8\cdot h - 8x-(-8x)}{h}\\&f'(x)=\frac{-8\cdot h}{h}\\&f'(x)=-8\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&-8\end{align*}$

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Example Question #23 : Concept Of The Derivative

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=-7x\text{ as h approaches 0.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=-7x\text{ and we can find that }\\&f(x+h)=- 7\cdot h - 7x\\&f'(x)=\frac{- 7\cdot h - 7x-(-7x)}{h}\\&f'(x)=\frac{-7\cdot h}{h}\\&f'(x)=-7\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&-7\end{align*}$

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Example Question #22 : Derivative Defined As The Limit Of The Difference Quotient

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=-3x^{3}\text{ as h approaches 0.}\end{align*}$

Possible Answers:

$3x^{2}$

$9x^{2}$

$-9x^{2}$

Correct answer:

$-9x^{2}$

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=-3x^{3}\text{ and we can find that }\\&f(x+h)=-3\cdot (h + x)^{3}\\&f'(x)=\frac{-3\cdot (h + x)^{3}-(-3x^{3})}{h}\\&f'(x)=\frac{- 9\cdot hx^{2} - 9\cdot h^{2}x - 3\cdot h^{3}}{h}\\&f'(x)=- 9\cdot hx - 3\cdot h^{2} - 9x^{2}\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&-9x^{2}\end{align*}$

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Example Question #21 : Derivative Defined As The Limit Of The Difference Quotient

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

Possible Answers:

18x

-18x

Correct answer:

-18x

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=-9x^{2}\text{ and we can find that }\\&f(x+h)=-9\cdot (h + x)^{2}\\&f'(x)=\frac{-9\cdot (h + x)^{2}-(-9x^{2})}{h}\\&f'(x)=\frac{- 18\cdot hx - 9\cdot h^{2}}{h}\\&f'(x)=- 9\cdot h - 18x\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&-18x\end{align*}$

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Example Question #24 : Derivative Defined As The Limit Of The Difference Quotient

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=2x - 2x^{2} - 2x^{3} - 1\text{ as h approaches 0.}\end{align*}$

Possible Answers:

$\text{Nonexistent.}$

$4x + 6x^{2} - 2$

$2 - 6x^{2} - 4x$

6x^2

Correct answer:

$2 - 6x^{2} - 4x$

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=2x - 2x^{2} - 2x^{3} - 1\text{ and we can find that }\\&f(x+h)=2\cdot h + 2x - 2\cdot (h + x)^{2} - 2\cdot (h + x)^{3} - 1\\&f'(x)=\frac{2\cdot h + 2x - 2\cdot (h + x)^{2} - 2\cdot (h + x)^{3} - 1-(2x - 2x^{2} - 2x^{3} - 1)}{h}\\&f'(x)=\frac{2\cdot h - 4\cdot hx - 6\cdot hx^{2} - 6\cdot h^{2}x - 2\cdot h^{2} - 2\cdot h^{3}}{h}\\&f'(x)=2 - 4x - 6\cdot hx - 2\cdot h^{2} - 6x^{2} - 2\cdot h\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&2 - 6x^{2} - 4x\end{align*}$

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Example Question #641 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=- 3x - 8x^{2} - 6\text{ as h approaches 0.}\end{align*}$

Possible Answers:

16x + 3

$\text{Nonexistent.}$

- 16x - 3

Correct answer:

- 16x - 3

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=- 3x - 8x^{2} - 6\text{ and we can find that }\\&f(x+h)=- 3\cdot h - 3x - 8\cdot (h + x)^{2} - 6\\&f'(x)=\frac{- 3\cdot h - 3x - 8\cdot (h + x)^{2} - 6-(- 3x - 8x^{2} - 6)}{h}\\&f'(x)=\frac{- 3\cdot h - 16\cdot hx - 8\cdot h^{2}}{h}\\&f'(x)=- 8\cdot h - 16x - 3\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&- 16x - 3\end{align*}$

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AP Calculus AB : Concept of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #21 : Concept Of The Derivative

Example Question #13 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #14 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #11 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #1181 : Ap Calculus Ab

Example Question #23 : Concept Of The Derivative

Example Question #22 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #21 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #24 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #641 : Derivatives

All AP Calculus AB Resources

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