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AP Calculus AB : Derivative defined as the limit of the difference quotient

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Example Question #641 : Derivatives

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

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 #36 : Concept Of The Derivative

Evaluate the following difference quotient:

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

Possible Answers:

-csc(x)cot(x)

-csc^2(x)

sec(x)tan(x)

sec(x)

-csc^2x

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 #31 : Derivative Defined As The Limit Of The Difference Quotient

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

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 #31 : Concept Of The Derivative

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

Possible Answers:

$\text{Nonexistent.}$

$\frac{75}{2}$

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 #39 : Concept Of The Derivative

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

Possible Answers:

$\text{Nonexistent.}$

$\frac{311}{8}$

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 #32 : Derivative Defined As The Limit Of The Difference Quotient

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

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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Example Question #1191 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=4x + 8x^{2} + 10\text{ as h approaches 0 for }x=13\end{align*}$

Possible Answers:

212

108

1414

$\frac{53}{2}$

Correct answer:

212

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=4x + 8x^{2} + 10\text{ and we can find that }\\&f(x+h)=4\cdot h + 4x + 8\cdot (h + x)^{2} + 10\\&f'(x)=\frac{4\cdot h + 4x + 8\cdot (h + x)^{2} + 10-(4x + 8x^{2} + 10)}{h}\\&f'(x)=\frac{4\cdot h + 16\cdot hx + 8\cdot h^{2}}{h}\\&f'(x)=8\cdot h + 16x + 4\\&\text{Then, if we take the limit as h goes to zero, we find at }x=13:\\&212\end{align*}$

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Example Question #1192 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=4 - 7x\text{ as h approaches 0 for }x=12\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=4 - 7x\text{ and we can find that }\\&f(x+h)=4 - 7x - 7\cdot h\\&f'(x)=\frac{4 - 7x - 7\cdot h-(4 - 7x)}{h}\\&f'(x)=\frac{-7\cdot h}{h}\\&f'(x)=-7\\&\text{Then, if we take the limit as h goes to zero, we find at }x=12:\\&-7\end{align*}$

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Example Question #1193 : Ap Calculus Ab

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=x - 7\text{ as h approaches 0 for }x=10\end{align*}$

Possible Answers:

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=x - 7\text{ and we can find that }\\&f(x+h)=h + x - 7\\&f'(x)=\frac{h + x - 7-(x - 7)}{h}\\&f'(x)=\frac{h}{h}\\&f'(x)=1\\&\text{Then, if we take the limit as h goes to zero, we find at }x=10:\\&1\end{align*}$

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Example Question #41 : Concept Of The Derivative

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=8x + 5x^{2} + 6\text{ as h approaches 0 for }x=0\end{align*}$

Possible Answers:

$\frac{8}{5}$

Correct answer:

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=8x + 5x^{2} + 6\text{ and we can find that }\\&f(x+h)=8\cdot h + 8x + 5\cdot (h + x)^{2} + 6\\&f'(x)=\frac{8\cdot h + 8x + 5\cdot (h + x)^{2} + 6-(8x + 5x^{2} + 6)}{h}\\&f'(x)=\frac{8\cdot h + 10\cdot hx + 5\cdot h^{2}}{h}\\&f'(x)=5\cdot h + 10x + 8\\&\text{Then, if we take the limit as h goes to zero, we find at }x=0:\\&8\end{align*}$

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AP Calculus AB : Derivative defined as the limit of the difference quotient

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #641 : Derivatives

Example Question #36 : Concept Of The Derivative

Example Question #31 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #31 : Concept Of The Derivative

Example Question #39 : Concept Of The Derivative

Example Question #32 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #1191 : Ap Calculus Ab

Example Question #1192 : Ap Calculus Ab

Example Question #1193 : Ap Calculus Ab

Example Question #41 : Concept Of The Derivative

All AP Calculus AB Resources

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