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AP Calculus AB : Concept of the Derivative

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Example Questions

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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Example Question #651 : Derivatives

Evaluate the following difference quotient:

$\lim_{x\rightarrow \Delta x}\frac{cot(x+\Delta x)-cot(x)}{\Delta x}$

Possible Answers:

-csc^2x

sec(x)

-csc(x)cot(x)

sec(x)tan(x)

-csc^2(x)

Correct answer:

-csc^2(x)

Explanation:

One should recognize the difference quotient as being equivalent to the derivative or instantaneous rate of change.

That is, for some continuous function, f(x) .

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

The function in our question is cot(x)

We note from this that because the difference quotient is the derivative:

$\lim_{x\rightarrow \Delta x}\frac{cot(x+\Delta x)-cot(x)}{\Delta x}=-csc^2(x)$

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Example Question #652 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=- x - 10\text{ as h approaches 0 for }x=-17\end{align*}$

Possible Answers:

$\text{Nonexistent.}$

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=- x - 10\text{ and we can find that }\\&f(x+h)=- h - x - 10\\&f'(x)=\frac{- h - x - 10-(- x - 10)}{h}\\&f'(x)=\frac{-h}{h}\\&f'(x)=-1\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-17:\\&-1\end{align*}$

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Example Question #653 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=4x - 8x^{2} + 8\text{ as h approaches 0 for }x=19\end{align*}$

Possible Answers:

$\frac{75}{2}$

$\text{Nonexistent.}$

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=4x - 8x^{2} + 8\text{ and we can find that }\\&f(x+h)=4\cdot h + 4x - 8\cdot (h + x)^{2} + 8\\&f'(x)=\frac{4\cdot h + 4x - 8\cdot (h + x)^{2} + 8-(4x - 8x^{2} + 8)}{h}\\&f'(x)=\frac{4\cdot h - 16\cdot hx - 8\cdot h^{2}}{h}\\&f'(x)=4 - 16x - 8\cdot h\\&\text{Then, if we take the limit as h goes to zero, we find at }x=19:\\&-300\end{align*}$

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Example Question #654 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=9 - 8x^{2} - 7x\text{ as h approaches 0 for }x=19\end{align*}$

Possible Answers:

$\frac{311}{8}$

$\text{Nonexistent.}$

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=9 - 8x^{2} - 7x\text{ and we can find that }\\&f(x+h)=9 - 7x - 8\cdot (h + x)^{2} - 7\cdot h\\&f'(x)=\frac{9 - 7x - 8\cdot (h + x)^{2} - 7\cdot h-(9 - 8x^{2} - 7x)}{h}\\&f'(x)=\frac{- 7\cdot h - 16\cdot hx - 8\cdot h^{2}}{h}\\&f'(x)=- 8\cdot h - 16x - 7\\&\text{Then, if we take the limit as h goes to zero, we find at }x=19:\\&-311\end{align*}$

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Example Question #655 : Derivatives

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=3x + x^{2} + 7\text{ as h approaches 0 for }x=-4\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=3x + x^{2} + 7\text{ and we can find that }\\&f(x+h)=3\cdot h + 3x + (h + x)^{2} + 7\\&f'(x)=\frac{3\cdot h + 3x + (h + x)^{2} + 7-(3x + x^{2} + 7)}{h}\\&f'(x)=\frac{3\cdot h + 2\cdot hx + h^{2}}{h}\\&f'(x)=h + 2x + 3\\&\text{Then, if we take the limit as h goes to zero, we find at }x=-4:\\&-5\end{align*}$

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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 #646 : Derivatives

Example Question #647 : Derivatives

Example Question #648 : Derivatives

Example Question #649 : Derivatives

Example Question #650 : Derivatives

Example Question #651 : Derivatives

Example Question #652 : Derivatives

Example Question #653 : Derivatives

Example Question #654 : Derivatives

Example Question #655 : Derivatives

All AP Calculus AB Resources

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