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AP Calculus BC : Computation of Derivatives

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Example Questions

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 #4 : Computation Of Derivatives

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

Possible Answers:

$e^{-x^2}$

$2xe^{-x^4}$

x^2

None of the other answers.

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 #378 : Ap Calculus Bc

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

Possible Answers:

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

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

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

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

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

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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Example Question #10 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

What is the rate of change of the function f(x)=1+2x+3x^4 at the point (1,6) ?

Possible Answers:

2594

Correct answer:

Explanation:

The rate of change of a function at a point is the value of the derivative at that point. First, take the derivative of f(x) using the power rule for each term.

Remember that the power rule is

$\frac{d}{dx}x^n=nx^{n-1}$ , and that the derivative of a constant is zero.

f'(x)=2+12x^3

Next, notice that the x-value of the point (1,6) is 1, so substitute 1 for x in the derivative.

f'(1)=2+12(1)^3=14

Therefore, the rate of change of f(x) at the point (1,6) is 14.

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AP Calculus BC : Computation of Derivatives

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

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 #4 : Computation Of Derivatives

Example Question #378 : Ap Calculus Bc

Example Question #379 : Ap Calculus Bc

Example Question #10 : Rules Of Basic Functions: Power, Exponential Rule, Logarithmic, Trigonometric, And Inverse Trigonometric

All AP Calculus BC Resources

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