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AP Calculus AB : AP Calculus AB

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Example Questions

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Example Question #63 : Calculus I — Derivatives

What is the second derivative of 12x^2+13x+4 ?

Possible Answers:

144

24x+13

24 x

Correct answer:

Explanation:

To find the second derivative, first we need to find the first derivative.

To do that, we can use the power rule. To use the power rule, we lower the exponent on the variable and multiply by that exponent.

We're going to treat as 4x^0 since anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})+(0*4x^{0-1})$

Notice that $(0*4x^{0-1})=0$ since anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{2-1})+(1*13x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(2*12x^{1})+(1*13x^{0})$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x^{1})+(13x^{0})$

Just like it was mentioned earlier, anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=(24x)+(13*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(12x^2+13x+4)=24x+13$

Now we repeat the process using 24x+13 as our expression.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(1*24x^{1-1})+(0*13x^{0-1})$

Like before, anything times zero is zero.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+10)=(1*24x^{1-1})$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24x^{0})$

Anything to the zero power is one.

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=(24*1)$

$\frac{\mathrm{d} }{\mathrm{d} x}(24x+13)=24$

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Example Question #51 : Calculus I — Derivatives

Define $g(x) = 7x^{5} - 6x^{3} - 17x$ .

What is g ''(x) ?

Possible Answers:

$g'' (x) = 140x^{3} - 18x$

$g'' (x) = 140x^{3} - 36x -17$

$g'' (x) = 140x^{3} - 36x$

$g'' (x) = 70x^{3} - 36x$

$g'' (x) = 70x^{3} - 18x$

Correct answer:

$g'' (x) = 140x^{3} - 36x$

Explanation:

Take the derivative of , then take the derivative of .

$g(x) = 7x^{5} - 6x^{3} - 17x$

$g' (x) = 5\cdot 7x^{5-1} - 3\cdot 6x^{3-1} - 17$

$g' (x) = 35x^{4} - 18x^{2} - 17$

$g'' (x) = 4\cdot 35x^{4-1} - 2\cdot 18x^{2-1} - 0$

$g'' (x) = 140x^{3} - 36x$

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Example Question #21 : Calculus I — Derivatives

Find $f^{'}(2)$ if the function f(x) is given by

f(x) = ln(x)

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

To find the derivative at x=2 , we first take the derivative of f(x) . By the derivative rule for logarithms,

$f'(x) = \frac{1}{x}$

Plugging in x=2 , we get

$f'(2) = \frac{1}{2}$

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Example Question #22 : Calculus I — Derivatives

Find the derivative of the following function at the point x=3 .

f(x) = 4x^3+x+3

Possible Answers:

106

108

109

110

107

Correct answer:

109

Explanation:

Use the power rule on each term of the polynomial to get the derivative,

f'(x) = (3)(4)x^2+1

Now we plug in x=3

f'(3) = (12)(3)^2 + 1 = 109

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Example Question #23 : Calculus I — Derivatives

Let $f(x)=x^2\sin(x^2)$ . What is $f'\left (\frac{\sqrt{\pi}}{2} \right )$ ?

Possible Answers:

$\frac{\sqrt{2\pi}(\pi + 4)}{8}$

$\frac{{2\pi}(\sqrt{\pi} + 2)}{4}$

$\frac{\sqrt{2\pi}(\pi + \sqrt{2})}{4}$

$\frac{(\pi + \sqrt{2})}{8}$

$\frac{\sqrt{2\pi}(\pi + 2)}{4}$

Correct answer:

$\frac{\sqrt{2\pi}(\pi + 4)}{8}$

Explanation:

We need to find the first derivative of f(x). This will require us to apply both the Product and Chain Rules. When we apply the Product Rule, we obtain:

$f'(x)=\sin(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]+x^2\cdot\frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]$

In order to find the derivative of $\sin(x^2)$ , we will need to employ the Chain Rule.

$\frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]=\cos(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]=\cos(x^2)\cdot2x$

$f'(x)=\sin(x^2)\cdot2x + x^2\cdot\cos(x^2)\cdot2x$

We can factor out a 2x to make this a little nicer to look at.

$f'(x)=2x(\sin(x^2)+x^2\cos(x^2))$

Now we must evaluate the derivative when x = $\frac{\sqrt{\pi}}{2}$ .

$f'\left (\frac{\sqrt{\pi}}{2} \right )=2\cdot\frac{\sqrt{\pi}}{2}(\sin\frac{\pi}{4}+\frac{\pi}{4}\cos\frac{\pi}{4})$

$=\sqrt{\pi}(\frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8})=\sqrt{2\pi}(\frac{1}{2}+\frac{\pi}{8})=\frac{\sqrt{2\pi}(\pi + 4)}{8}$

The answer is $\frac{\sqrt{2\pi}(\pi + 4)}{8}$ .

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Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

Find the derivative of the following function:

$f(x) = x^3 + \frac{1}{x}$

Possible Answers:

$f'(x) = 3x^2-\frac{1}{x}$

f'(x) = 3x^2+x

$f'(x) = 3x^2+\frac{1}{x^2}$

f'(x) = 3x^2

$f'(x) = 3x^2-\frac{1}{x^2}$

Correct answer:

$f'(x) = 3x^2-\frac{1}{x^2}$

Explanation:

Since this function is a polynomial, we take the derivative of each term separately.

From the power rule, the derivative of

x^3

is simply

3x^2

We can rewrite $\frac{1}{x}$ as

$x^{-1}$

and using the power rule again, we get a derivative of

$-x^{-2}$ or $-\frac{1}{x^2}$

So the answer is

$f'(x) = 3x^2-\frac{1}{x^2}$

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Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

What is $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} ?$

Possible Answers:

f(x)*f'(x)+g(x)*g'(x)

f(x)g'(x)-f'(x)g(x)

f(x)g(x)+f'(x)g'(x)

$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

f(x)g'(x)+f'(x)g(x)

Correct answer:

f(x)g'(x)+f'(x)g(x)

Explanation:

The chain rule is "first times the derivative of the second plus second times derivative of the first".

In this case, that means $\frac{\mathrm{d} }{\mathrm{d} x}f(x)*g(x)\mathrm{d x} =f(x)g'(x)+f'(x)g(x)$ .

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Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

Which of the following best represents $\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$ ?

Possible Answers:

$\frac{f(x)g'(x)-g(x)f'(x)}{g(x)^2}$

$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

$\frac{g(x)g'(x)-f(x)f'(x)}{g(x)^2}$

f'(x)g(x)+g'(x)f(x)

$\frac{f'(x)}{g'(x)}$

Correct answer:

$\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$

Explanation:

The question is just asking for the Quotient Rule formula.

Recall the Quotient Rule is the bottom function times the derivative of the top minus the top function times the derivative of the bottom all divided by the bottom function squared.

Given,

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}$

the bottom function is g(x) and the top function is f(x) . This makes the bottom derivative g'(x) and the top derivative f'(x) .

Substituting these into the Quotient Rule formula resulting in the following.

$\frac{\mathrm{d} }{\mathrm{d} x}\frac{f(x)}{g(x)}=\frac{g(x)\cdot f'(x)-f(x)\cdot g'(x)}{(g(x))^2}$

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Example Question #1 : Specific Derivatives

Find the derivative $y^{'}$ for $y = x^{2}2^x$

Possible Answers:

$2x2^{x}-x^22^{x-1}$

$x2^{x+1}+\ln(2)x^{2}2^{x}$

$2x2^{x}+x^{3}2^{x-1}$

$2x\ln(2)2^{x}$

Correct answer:

$x2^{x+1}+\ln(2)x^{2}2^{x}$

Explanation:

The derivative must be computed using the product rule. Because the derivative of x^2 brings a down as a coefficient, it can be combined with 2^x to give $2^{x+1}$

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Example Question #3 : Specific Derivatives

$f(x) = 5 ^{x + 1}$

What is f'(x) ?

Possible Answers:

$f'(x) = \frac{5 ^{x}}{\ln 5}$

$f'(x) = 5 ^{x}$

$f'(x) = \ln 5 \cdot 5 ^{x+1}$

$f'(x) = \frac{5 ^{x+1}}{\ln 5}$

$f'(x) = \ln 5 \cdot 5 ^{x}$

Correct answer:

$f'(x) = \ln 5 \cdot 5 ^{x+1}$

Explanation:

$f(x) = 5 ^{x + 1} = 5 ^{1} \cdot 5 ^{x } = 5 \cdot 5 ^{x }$

Therefore,

$f'(x) = 5 \cdot \frac{\mathrm{d} }{\mathrm{d} x} \left ( 5 ^{x } \right )$

$\frac{\mathrm{d}}{\mathrm{d} x} \left ( a ^{x } \right ) = \ln a \cdot a ^{x }$ for any real , so $\frac{\mathrm{d} }{\mathrm{d} x} \left ( 5 ^{x } \right ) = \ln 5 \cdot 5 ^{x}$ , and

$f'(x) = 5 \ln 5 \cdot 5 ^{x} = \ln 5 \cdot 5 \cdot 5 ^{x} = \ln 5 \cdot 5 ^{x+1}$

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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 #63 : Calculus I — Derivatives

Example Question #51 : Calculus I — Derivatives

Example Question #21 : Calculus I — Derivatives

Example Question #22 : Calculus I — Derivatives

Example Question #23 : Calculus I — Derivatives

Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

Example Question #1 : Understanding The Derivative Of A Sum, Product, Or Quotient

Example Question #1 : Specific Derivatives

Example Question #3 : Specific Derivatives

All AP Calculus AB Resources

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