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AP Calculus AB : Derivatives

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Example Questions

Example Question #641 : Derivatives

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

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 #642 : Derivatives

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

Possible Answers:

$-9x^{2}$

$9x^{2}$

$3x^{2}$

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 #643 : Derivatives

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

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 #644 : Derivatives

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

Possible Answers:

6x^2

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

$\text{Nonexistent.}$

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

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 #645 : 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:

$\text{Nonexistent.}$

- 16x - 3

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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Example Question #646 : Derivatives

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

Possible Answers:

$\text{Nonexistent.}$

Correct answer:

Explanation:

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

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Example Question #647 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=3x^{2} - 5x - 1\text{ as h approaches 0.}\end{align*}$

Possible Answers:

$\text{Nonexistent.}$

6x - 5

5 - 6x

Correct answer:

6x - 5

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=3x^{2} - 5x - 1\text{ and we can find that }\\&f(x+h)=3\cdot (h + x)^{2} - 5x - 5\cdot h - 1\\&f'(x)=\frac{3\cdot (h + x)^{2} - 5x - 5\cdot h - 1-(3x^{2} - 5x - 1)}{h}\\&f'(x)=\frac{6\cdot hx - 5\cdot h + 3\cdot h^{2}}{h}\\&f'(x)=3\cdot h + 6x - 5\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&6x - 5\end{align*}$

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Example Question #648 : Derivatives

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

Possible Answers:

Correct answer:

Explanation:

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

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Example Question #649 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=7 - 8x\text{ as h approaches 0.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=7 - 8x\text{ and we can find that }\\&f(x+h)=7 - 8x - 8\cdot h\\&f'(x)=\frac{7 - 8x - 8\cdot h-(7 - 8x)}{h}\\&f'(x)=\frac{-8\cdot h}{h}\\&f'(x)=-8\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&-8\end{align*}$

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Example Question #650 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=- 5x - 1\text{ as h approaches 0.}\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=- 5x - 1\text{ and we can find that }\\&f(x+h)=- 5\cdot h - 5x - 1\\&f'(x)=\frac{- 5\cdot h - 5x - 1-(- 5x - 1)}{h}\\&f'(x)=\frac{-5\cdot h}{h}\\&f'(x)=-5\\&\text{Then, if we take the limit as h goes to zero, we find:}\\&-5\end{align*}$

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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 #641 : Derivatives

Example Question #642 : Derivatives

Example Question #643 : Derivatives

Example Question #644 : Derivatives

Example Question #645 : Derivatives

Example Question #646 : Derivatives

Example Question #647 : Derivatives

Example Question #648 : Derivatives

Example Question #649 : Derivatives

Example Question #650 : Derivatives

All AP Calculus AB Resources

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