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AP Calculus AB : Concept of the Derivative

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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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Example Question #41 : Derivative Defined As The Limit Of The Difference Quotient

$\begin{align*}&\text{Calculate the limit of the difference quotient }f'(x)=\frac{f(x+h)-f(x)}{h}\\&\text{For }f(x)=x + 9x^{2} - 6\text{ as h approaches 0 for }x=8\end{align*}$

Possible Answers:

217

578

145

Correct answer:

145

Explanation:

$\begin{align*}&\text{The difference quotient is a means of approximating a rate of}\\&\text{change over a given increment, h. As the increment, h, grows}\\&\text{smaller and smaller, we get increasingly refine this approximation,}\\&\text{gaining an increasingly accurate representation of an instantaneous}\\&\text{rate of change, rather than a broader average. This is what}\\&\text{is also known as the derivative. For our problem:}\\&f(x)=x + 9x^{2} - 6\text{ and we can find that }\\&f(x+h)=h + x + 9\cdot (h + x)^{2} - 6\\&f'(x)=\frac{h + x + 9\cdot (h + x)^{2} - 6-(x + 9x^{2} - 6)}{h}\\&f'(x)=\frac{h + 18\cdot hx + 9\cdot h^{2}}{h}\\&f'(x)=9\cdot h + 18x + 1\\&\text{Then, if we take the limit as h goes to zero, we find at }x=8:\\&145\end{align*}$

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Example Question #42 : Derivative Defined As The Limit Of The Difference Quotient

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

Possible Answers:

$-\frac{59}{3}$

590

580

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

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

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

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

Possible Answers:

$\text{Nonexistent.}$

401

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

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

f(x)=12x+3

Possible Answers:

None of the other answers

The limit does not exist

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

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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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 #1191 : Ap Calculus Ab

Example Question #1192 : Ap Calculus Ab

Example Question #1193 : Ap Calculus Ab

Example Question #41 : Concept Of The Derivative

Example Question #41 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #42 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #43 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #41 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #45 : Derivative Defined As The Limit Of The Difference Quotient

Example Question #46 : Derivative Defined As The Limit Of The Difference Quotient

All AP Calculus AB Resources

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