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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

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Example Questions

Example Question #371 : Ap Calculus Bc

Assuming f(x) is continuous and differentiable for all values of x, what can be said about its graph at the point x=5 if we know that ?

Possible Answers:

f(x) is concave up when x=5

f(x) is decreasing when x=5

There is not sufficient information to describe f(x).

f(x) is increasing when x=5

Correct answer:

f(x) is increasing when x=5

Explanation:

Assuming f(x) is continuous and differentiable for all values of x, what can be said about its graph at the point x=5 if we know that ?

We are told about a first derivative and asked to consider the original function. Recall that anywhere the first derivative is negative, the original function is decreasing. Anywhere the first derivative is positive, the original function is increasing.

We are told that . In other words, the first derivative is positive.

This means our original function must be increasing.

f(x) is increasing when x=5

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Example Question #372 : Ap Calculus Bc

$f (x) = 7x^{2} + 6x - 9$

Give f'(x) .

Possible Answers:

$f'(x) = 14x^{2}+6x$

f'(x) = 14x+6

f'(x) = 7x+6

f'(x) = 14x

f'(x) = 14x+12

Correct answer:

f'(x) = 14x+6

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f (x) = 7x^{2} + 6x - 9$

$f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0$

$f ' (x) = 14\cdot x^{1} + 6$

f ' (x) = 14x + 6

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Example Question #373 : Ap Calculus Bc

$f(x)= 8x^{3}-7x^{2}+11$

Give f''(x) .

Possible Answers:

f ''(x) = 48x -14

f ''(x) = 24x

f ''(x) = 24x -14

f ''(x) = 24x - 11

f ''(x) = 48x

Correct answer:

f ''(x) = 48x -14

Explanation:

First, find the derivative f'(x) of $f(x)= 8x^{3}-7x^{2}+11$ .

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 8x^{3}-7x^{2}+11$

$f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0$

$f'(x)= 24x^{2}-14x^{1}$

$f'(x)= 24x^{2}-14x$

Now, differentiate f'(x) to get f''(x) .

$f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}$

$f''(x)= 48x^{1}-14x^{0}$

f''(x)= 48x-14

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Example Question #374 : Ap Calculus Bc

Differentiate $x^{999}$ .

Possible Answers:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{999}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 1,000x^{1,000}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{998}$

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 998x^{1,000}$

Correct answer:

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{998}$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}$

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Example Question #375 : Ap Calculus Bc

Give the second derivative of $x^{777}$ .

Possible Answers:

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{779}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 604,506 x^{779}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 603,729 x^{777}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 603,729 x^{775}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{775}$

Correct answer:

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = 602,952 x^{775}$

Explanation:

Find the derivative of $x^{777}$ , then find the derivative of that expression.

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , so

$\frac{\mathrm{d} }{\mathrm{d} x} \left (x^{777} \right )= 777\cdot x^{777-1} = 777x^{776}$

$\frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 777x^{776} \right )= 777\cdot776\cdot x^{776-1} = 602,952x^{775}$

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Example Question #376 : Ap Calculus Bc

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

Give f'(x) .

Possible Answers:

$f'(x) = 2.4x^{3}-x^{2}$

$f'(x) = 2.4x^{2}-x$

$f'(x) = 2.4x^{3}-x^{2}+ 0.9$

$f'(x) = 2.4x^{2}-x + 0.9$

$f'(x) = 2.4x^{2}-1$

Correct answer:

$f'(x) = 2.4x^{2}-x$

Explanation:

$\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0, so

$f(x)= 0.8x^{3}-0.5x^{2}+ 0.9$

$f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0$

$f'(x)= 2.4x^{2}-1x^{1}$

$f'(x) = 2.4x^{2}-x$

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Example Question #377 : Ap Calculus Bc

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

Give f''(x) .

Possible Answers:

$f''(x) =10.8x^{2} + 4.2x$

$f''(x) =10.8x^{2} + 8.4x$

$f''(x) =10.8x^{2} + 4.2x - 0.5$

$f''(x) =10.8x^{2} + 2.1x$

$f''(x) =10.8x^{2} + 8.4x - 0.5$

Correct answer:

$f''(x) =10.8x^{2} + 4.2x$

Explanation:

First, find the derivative f'(x) of $f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$ .

Recall that $\frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}$ , and the derivative of a constant is 0.

$f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x$

$f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}$

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

Now, differentiate f'(x) to get f''(x) .

$f'(x) =3.6x^{3} + 2.1x^{2}- 0.5$

$f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0$

$f''(x) =10.8x^{2} + 4.2x^{1}- 0$

$f''(x) =10.8x^{2} + 4.2x$

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Example Question #1 : Computation Of Derivatives

Find the derivative of the function $y = \int_{0}^{x^2} e^{-t^2}dt$

Possible Answers:

None of the other answers.

$2xe^{-x^4}$

$e^{-x^2}$

x^2

Correct answer:

$2xe^{-x^4}$

Explanation:

We can use the (first part of) the Fundemental Theorem of Calculus to "cancel out" the integral.

$y = \int_{0}^{x^2} e^{-t^2}dt$ . Start

$y' = \frac{d}{dx}\int_{0}^{x^2} e^{-t^2}dt$ . Take the derivative of both sides with respect to .

To "cancel out" the integral and the derivative sign, verify that the lower bound on the integral is a constant (It's in this case), and that the upper limit of the integral is a function of , (it's x^2 in this case).

Afterward, plug x^2 in for , and ultilize the Chain Rule to complete using the Fundemental Theorem of Calculus.

$y' = e^{-(x^2)^2}(x^2)' = 2xe^{-x^4}$ .

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Example Question #1 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Find the derivative of: y=cos^-^1(x)+cos(x)

Possible Answers:

$\frac{1-sin(x)}{\sqrt{1+x^2}}$

$-\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

$\frac{1+cos(x)}{\sqrt{1+x^2}}$

$\frac{cos(x)\sqrt{1-x^2}-sin(x)}{\sqrt{1-x^2}}$

$-\frac{1+sin(x)\sqrt{1+x^2}}{\sqrt{1+x^2}}$

Correct answer:

$-\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

Explanation:

The derivative of inverse cosine is:

$\frac{d}{dx} cos^-^1(x) = -\frac{1}{\sqrt{1-x^2}}$

The derivative of cosine is:

$\frac{d}{dx} cos(x)=-sin(x)$

Combine the two terms into one term.

$-\frac{1}{\sqrt{1-x^2}}-sin(x)$

$-\frac{1}{\sqrt{1-x^2}}-\frac{sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}= -\frac{1+sin(x)\sqrt{1-x^2}}{\sqrt{1-x^2}}$

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Example Question #379 : Ap Calculus Bc

Find the derivative of the function $y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$

Possible Answers:

$x^2e^x\ln(x)\sin(x)$

Does not exist

None of the other answers

$\frac{2xe^x\cos{x}}{x}$

Correct answer:

$x^2e^x\ln(x)\sin(x)$

Explanation:

To find the derivative of this function, we need to use the Fundemental Theorem of Calculus Part 1 (As opposed to the 2nd part, which is what's usually used to evaluate definite integrals)

$y= \int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Start

$y'= \frac{d}{dx}\int_{1}^{x}t^2e^t\ln(t)\sin(t)dt$ . Take derivatives of both sides.

$y'= x^2e^x\ln(x)\sin(x)$ . "Cancel" the integral and the derivative. (Make sure that the upper bound on the integral is a function of , and that the lower bound is a constant before you cancel, otherwise you may need to use some manipulation of the bounds to make it so.)

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AP Calculus BC : AP Calculus BC

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #371 : Ap Calculus Bc

Example Question #372 : Ap Calculus Bc

Example Question #373 : Ap Calculus Bc

Example Question #374 : Ap Calculus Bc

Example Question #375 : Ap Calculus Bc

Example Question #376 : Ap Calculus Bc

Example Question #377 : Ap Calculus Bc

Example Question #1 : Computation Of Derivatives

Example Question #1 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

Example Question #379 : Ap Calculus Bc

All AP Calculus BC Resources

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