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AP Calculus AB : AP Calculus AB

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Example Questions

Example Question #1201 : Ap Calculus Ab

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

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

145

578

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 #41 : Concept Of The Derivative

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

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 #1204 : Ap Calculus Ab

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

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 #1205 : Ap Calculus Ab

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

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 #1206 : Ap Calculus Ab

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

f(x)=12x+3

Possible Answers:

The limit does not exist

None of the other answers

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 #41 : 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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Example Question #51 : Concept Of The Derivative

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

f(x)=5x^2+1

Possible Answers:

10x+5h

10x

15xh

10x+1

Correct answer:

10x

Explanation:

The derivative of a function f(x) is defined by the limit of the difference quotient, as follows:

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

Using this limit for our function, and remembering to write the limit at every step, we get

$\lim_{h\rightarrow 0} \frac{5(x+h^2)+1-(5x^2+1)}{h}=\lim_{h\rightarrow 0} \frac{5(x^2+2xh+h^2)+1-5x^2-1}{h}=\lim_{h\rightarrow 0}\frac{10xh+5h^2}{h}=\lim_{h\rightarrow 0} (10x+5h) = 10x$

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Example Question #54 : Derivatives

If y= ln(x^3) what is the slope of the line at x=12 .

Possible Answers:

0.25

0.5

0.1

$\infty$

Correct answer:

0.25

Explanation:

The slope at any point on a line is also equal to the derivative. So first we want to find the derivative function of this function and then evaluate it at x=12 . So, to find the derivative we will need to use the chain rule. The chain rule says

$\frac{d}{dx}} f(g(x)) = f'(g(x))g'(x)$

so if we let f(x)= ln(x) and g(x)=x^3 then

since $f'(x)=\frac{1}{x}}$ and g'(x)=3x^2

$\frac{d}{dx}} f(g(x)) = f'(g(x))g'(x) = \frac{1}{x^3}*3x^2 = \frac{3}{x}$

Therefore we evaluate at x=12 and we get $\frac{3}{12}$ or .

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Example Question #2 : Using The Chain Rule

What is the first derivative of (x+6)^2 ?

Possible Answers:

2x+12

2x+13

x+6

4x+24

Correct answer:

2x+12

Explanation:

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: $\frac{\mathrm{d} }{\mathrm{d} x}f(g(x))=f'(g(x))*g'(x)$

Plug in our equations.

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6)^{1})*(1+0)$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=(2(x+6))$

$\frac{\mathrm{d} }{\mathrm{d} x}(x+6)^2=2x+12$

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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #1201 : Ap Calculus Ab

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

Example Question #41 : Concept Of The Derivative

Example Question #1204 : Ap Calculus Ab

Example Question #1205 : Ap Calculus Ab

Example Question #1206 : Ap Calculus Ab

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

Example Question #51 : Concept Of The Derivative

Example Question #54 : Derivatives

Example Question #2 : Using The Chain Rule

All AP Calculus AB Resources

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