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AP Calculus AB : AP Calculus AB

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3 Diagnostic Tests 164 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #85 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of the function:

$h(g)=e^{cos(g)}$

Possible Answers:

$h'(g)=sin(g)e^{cos(g)}$

$h'(g)=2sin(g)e^{2sin(g)}$

$h'(g)=sin(g)e^{sin(g)}$

$h'(g)=-2sin(g)e^{2sin(g)}$

$h'(g)=-sin(g)e^{cos(g)}$

Correct answer:

$h'(g)=-sin(g)e^{cos(g)}$

Explanation:

*****************************************************************

STEPS

*****************************************************************

Given:

$h(g)=e^{cos(g)}$

Understand the derivative of the exponential:

$h'(g)=e^{cos(g)}{\color{Blue} *(chain)}$

To obtain the chained term, we must derive the compound function element, in this case, the cos function contained in the exponent

${\color{Blue} d(cos(g))=-sin(g)}$

Plug the -sin back in:

$h'(g)=e^{cos(g)}{\color{Blue} *-sin(g)}$

Pull the sin to the front, and we arrive at the correct answer:

*****************************************************************

ANSWER

*****************************************************************

${\color{Red} h'(g)=-sin(g)e^{cos(g)}}$

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Example Question #86 : Derivative Rules For Sums, Products, And Quotients Of Functions

Calculate the derivative of $\frac{ln(x)}{x}$ .

Possible Answers:

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

$\frac{1-ln(x)}{x}$

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

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

Correct answer:

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

Explanation:

We don't have a formula for taking the derivative of this expression, so we'll have to use the quotient rule, since we have a fraction of functions.

The quotient rule is $(\frac{f}{g})'=\frac{g'f-f'g}{g^2}$ .

Applying it to $\frac{ln(x)}{x}$ , we get

$\frac{x*\tfrac{1}{x}-ln(x)*1}{x^2}=\frac{1-ln(x)}{x^2}$

Our answer is $\frac{1-ln(x)}{x^2}$ .

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Example Question #87 : Derivative Rules For Sums, Products, And Quotients Of Functions

Find the derivative of $x^2\ln x$ .

Possible Answers:

$x+\ln x$

$x+x\ln x$

$\ln x$

$x+2x\ln x$

$2x\ln x$

Correct answer:

$x+2x\ln x$

Explanation:

Using the product rule, (fg)'=f*g'+g*f' , the derivative of $x^2\ln x$ is

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

Final answer:

$x+2x\ln x$

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Example Question #291 : Computation Of The Derivative

Find the derivative of $15x^5-2x^4+11x^2+e^x-\ln x$ .

Possible Answers:

$15x^4-2x^3+11x+e^x-\ln x$

$25x^4+2x^3+11x+e^x+\ln x$

$75x^4+8x^3-22x-e^x+\tfrac{1} {x}$

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

55x^4-6x^3+22x

Correct answer:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

Explanation:

Take the derivative of each term.

$\tfrac{\mathrm{d} }{\mathrm{d} x}(15x^5)=75x^4$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-2x^4)=-8x^3$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(11x^2)=22x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(e^x)=e^x$

$\tfrac{\mathrm{d} }{\mathrm{d} x}(-\ln x)=-\tfrac{1}{x}$

Add them:

$75x^4-8x^3+22x+e^x-\tfrac{1} {x}$

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Example Question #292 : Computation Of The Derivative

Find the derivative of $2\sin x\cos x$ .

Possible Answers:

$-2\cos x\sin x$

$2\sin2x$

$2\cos2x$

$-\sin^2x+cos^2x$

Correct answer:

$2\cos2x$

Explanation:

There are two ways to solve this problem.

First, you can use a trig identity to replace $2\sin x\cos x$ with $\sin 2x$ . Using the chain rule, $(\sin 2x)'=2\cos2x$ .

Alternatively, you could use the product rule.

$2(\sin x(-\sin x)+\cos x(\cos x))=2(\cos^2x-\sin^2x)$

Since $\cos^2x-\sin^2x=\cos2x$ , our final answer is still $2\cos2x$ .

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Example Question #293 : Computation Of The Derivative

Find the derivative of $\frac{e^x}{\sin x}$ .

Possible Answers:

$\frac{-e^x\tan x}{\sin x}$

$\frac{e^x\tan x-e^x}{\cos x}$

$\frac{e^x-e^x\tan x}{\sin x}$

$\frac{e^x\sin x-e^x\cos x}{\sin x}$

$\frac{e^x}{\cos x}$

Correct answer:

$\frac{e^x-e^x\tan x}{\sin x}$

Explanation:

For this problem, we need to use the quotient rule.

$(\frac{f}{g})'=\frac{gf'-fg'}{g^2}$

$(\frac{e^x}{\sin x})'=\frac{\sin x(e^x)-e^x(\cos x)}{\sin^2x}$

Simplifying:

$e^x (\frac{\sin x-\cos x}{\sin^2x})=\frac{e^x-e^x\tan x}{\sin x}$

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Example Question #461 : Derivatives

Find the derivative of y=(2x^4+5x)(cos(x))

Possible Answers:

y'=8x^3cos(x)+5cos(x)

y'=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

y'=2x^4cos(x)-2x^4sin(x)+5xcos(x)-5xsin(x)

None of the other answers

Correct answer:

y'=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

Explanation:

$\frac{d}{dx}[2x^4+5x]=8x^3+5$

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

Product rule states: $\frac{d}{du}[f(u) \cdot g(u)]=f'(u)g(u)+g'(u)f(u)$

Therefore:

y'=(8x^3+5)(cos(x))-sin(x)(2x^4+5x)

=-2x^4sin(x)+8x^3cos(x)-5xsin(x)+5cos(x)

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Example Question #294 : Computation Of The Derivative

Use the chain rule to differentiate the following function: $h(x)=sin(x^{2}e^{x})$

Possible Answers:

$h^{'}(x)=-sin(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=-cos(x^{2}e^{x})xe^{x}(2+x)$

$h^{'}(x)=sin(x^{2}e^{x})xe^{x}(2+x)$

Correct answer:

$h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

Explanation:

$h(x)=sin(x^{2}e^{x})$

By the chain rule: $h^{'}(x)=cos(x^{2}e^{x})\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$

Differentiate $(x^{2}e^{x})$ using the product rule: $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})=\frac{\mathrm{d} }{\mathrm{d} x}(x^{2})e^{x}+x^{2}\frac{\mathrm{d} }{\mathrm{d} x}e^{x}=2xe^{x}+x^{2}e^{x}$

Substitute this derivative for $\frac{\mathrm{d} }{\mathrm{d} x}(x^{2}e^{x})$ in the first equation: $h^{'}(x)=cos(x^{2}e^{x})(2xe^{x}+x^{2}e^{x})$

Factor the equation: $h^{'}(x)=cos(x^{2}e^{x})xe^{x}(2+x)$

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Example Question #295 : Computation Of The Derivative

f(p)=cos(2p-1)-(2p^3+4p)

Find f'(p) .

Possible Answers:

f'(p)=cos(2p-1)+sin(2p-1)-6p^2+4

f'(p)=-2cos(2p-1)(p^3+3p^2+2p+2)

f'(p) is undefined.

f'(p)=sin(2p-1)-sin(2p^3+4p)

f'(p)=-2(sin(2p-1)+3p^2+2)

Correct answer:

f'(p)=-2(sin(2p-1)+3p^2+2)

Explanation:

$\frac{d}{dx}[cos(2p-1)]=2 \cdot -sin(2p-1)=-2sin(2p-1)$

$\frac{d}{dx}[2p^3+4p]=6p^2+4$

Therefore:

f'(p)=-2sin(2p-1)-(6p^2+4)

=-2sin(2p-1)-6p^2-4

=-2(sin(2p-1)+3p^2+2)

f'(p)=-2(3p^2+sin(2p-1)+2)

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Example Question #1 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

Use the fundamental theorem of Calculus to evaluate the definite integral

$\int_{\pi/2}^{\pi}(2cos(x)+1)dx$

Possible Answers:

$2-\frac{\pi]}{2}$

$\frac{\pi}{2}-2$

$2+\pi$

$\pi-2$

Correct answer:

$\frac{\pi}{2}-2$

Explanation:

Here we use the fundamental theorem of Calculus: $\int_{a}^{b}f(x)dx=F(b)-F(a)$

$F(x)=\int(2cos(x)+1)dx=2sin(x)+x$

Here we do not worry about adding a constant c because we are evaluating a definite integral.

$F(b)=F(\pi)=2sin(\pi)+\pi=0+\pi=\pi$

$F(a)=F(\frac{\pi}{2})=2sin(\frac{\pi}{2})+\frac{\pi}{2}= 2+\frac{\pi}{2}$

$F(b)-F(a)=F(\pi)-F(\frac{\pi}{2})=\pi-(2+\frac{\pi}{2})=\frac{\pi}{2}-2$

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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 #85 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #86 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #87 : Derivative Rules For Sums, Products, And Quotients Of Functions

Example Question #291 : Computation Of The Derivative

Example Question #292 : Computation Of The Derivative

Example Question #293 : Computation Of The Derivative

Example Question #461 : Derivatives

Example Question #294 : Computation Of The Derivative

Example Question #295 : Computation Of The Derivative

Example Question #1 : Use Of The Fundamental Theorem To Evaluate Definite Integrals

All AP Calculus AB Resources

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