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AP Calculus AB : AP Calculus AB

Study concepts, example questions & explanations for 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 #34 : Chain Rule And Implicit Differentiation

Find the derivative of:

$h(t)=\ln(\ln t)$

Possible Answers:

$h'(t)=\frac{\ln t}{t}$

$h'(t)=\frac{t}{\ln t}$

$h'(t)=t\ln t$

$h'(t)=\frac{1}{t\ln t}$

$h'(t)=\frac{1}{\ln t}+t$

Correct answer:

$h'(t)=\frac{1}{t\ln t}$

Explanation:

On this problem we have to use chain rule, which is:

$\frac{d}{dt}g(f(t))=g'(f(t))f'(t)$

So in this problem we let

$g(t)=\ln(t)$

and

$f(t)=\ln(t)$ .

Since

$g'(t)=\frac{1}{t}$

and

$f'(t)=\frac{1}{t}$ ,

we can conclude that

$h'(t)=\frac{d}{dt}g(f(t))=\frac{1}{f(t)}\cdot\frac{1}{t}=\frac{1}{t\ln t}$

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Example Question #35 : Chain Rule And Implicit Differentiation

Find the derivative of:

$f(x)=\ln(\sin e^x)$

Possible Answers:

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

$f'(x)=e^x\tan e^x$

$f'(x)=e^x\cot e^x$

$f'(x)=e^x\cos^2 e^x$

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

Correct answer:

$f'(x)=e^x\cot e^x$

Explanation:

On this problem we have to use chain rule, which is:

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

So in this problem we let

$g(x)=\ln x$

and

$h(x)=\sin x$

and

j(x)=e^x .

Since

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

and

$h'(x)=\cos x$

and

j'(x)=e^x ,

we can conclude that

$f'(x)=\frac{d}{dx}g(h(j(x)))=\frac{1}{h(j(x))}\cdot cos(j(x))\cdot e^x$

$f'(x)=\frac{1}{\sin e^x}\cdot \cos e^x\cdot e^x=e^x\cot e^x$

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Example Question #36 : Chain Rule And Implicit Differentiation

Find dy/dx of:

$y=e^{\cos x}$

Possible Answers:

$\frac{dy}{dx}= -e^{\cos x}\sin x$

$\frac{dy}{dx}= e^{\cos x}\cos x$

$\frac{dy}{dx}= =e^{\sin x}\cos x$

$\frac{dy}{dx}= e^{\cos x}\sin x$

$\frac{dy}{dx}= -e^{\cos x}\ln(\cos x)\cdot\sin x$

Correct answer:

$\frac{dy}{dx}= -e^{\cos x}\sin x$

Explanation:

On this problem we have to use chain rule, which is:

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

So in this problem we let

g(x)=e^x

and

$f(x)=\cos x$ .

Since

g'(x)=e^x

and

$f'(x)=-\sin x$ ,

we can conclude that

$\frac{dy}{dx}=\frac{d}{dx}g(f(x))=e^{(f(x))}(-\sin x) = -e^{\cos x}\sin x$

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Example Question #37 : Chain Rule And Implicit Differentiation

Find the derivative of the following equation:

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

Possible Answers:

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

$2x\cos(2x)$

$\cos(x^{2})$

None of the other answers

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

Correct answer:

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

Explanation:

Because we are differentiating a function within another function, we must use the chain rule, which gives that

(f o g(x)){}'=g{}'(x)f{}'(g(x)) .

Using this rule, we see that

$f(x)=\sin(x)$ ,

$g(x)=x^{2}$ ,

and therefore, the differentiation of

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

$2x\cos(x^{2})$ .

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Example Question #38 : Chain Rule And Implicit Differentiation

Find the derivative of the following equation:

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

Possible Answers:

$e^{2x}$

$2xe^{2x}$

None of the other answers.

$2e^{2x}$

$2xe^{2x-1}$

Correct answer:

$2e^{2x}$

Explanation:

Because we are differentiating a function within another function, we must use the chain rule, which states that

Looking at our function, we should be able to tell that

$f(x)=e^{x}$

and

g(x)=2x .

Given this, we can use the chain rule to solve:

$\frac{d}{dx}(e^{2x})=2e^{2x}$ .

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Example Question #39 : Chain Rule And Implicit Differentiation

Find the derivative of the following function:

$h(x)=5e^{\sin (x)}$

Possible Answers:

$\sin (x)e^{\sin (x)-1}$

$5\cos (x)e^{\sin (x)}$

$-5\cos (x)e^{\sin (x)}$

None of the other answers.

$5e^{\sin (x)}$

Correct answer:

$5\cos (x)e^{\sin (x)}$

Explanation:

Because we are differentiating a function within another function, we must use the chain rule, which states that

By examining the given equation

$h(x)=5e^{\sin (x)}$ ,

we see that we can find the derivative by pulling out the 5, as it is simply a constant:

$h{}'(x)=5[e^{\sin (x)}\frac{d}{dx}]$ .

We can see from this that

$f(x)=e^{x}$

and

$g(x)=\sin (x)$ .

By plugging this information into the chain rule, we find that the derivative is

$5[e^{\sin (x)}\cos (x)]=5\cos (x)e^{\sin (x)}$ .

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Example Question #40 : Chain Rule And Implicit Differentiation

Find the derivative of the following equation:

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

Possible Answers:

$2x^{3}e^{x^{2}}$

$2xe^{x^{2}}$

$4xe^{x^{2}}$

$2xe^{2x}$

$x^{2}e^{2x}$

Correct answer:

$2xe^{x^{2}}$

Explanation:

Because we are differentiating a function within another function, we must use the chain rule, which states that

from the given equation,

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

we can deduce that in this case,

$f(x)=e^{x}$ and $g(x)=x^{2}$ .

By plugging this into the chain rule, we find that

$h{}'(x)=2xe^{x^{2}}$ .

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Example Question #41 : Chain Rule And Implicit Differentiation

Find the derivative of the following equation:

$h(x)=\tan(\sin (x))$

Possible Answers:

$\sin(x)\sec ^{2}(\sin (x))$

$\sec^{2}(\sin (x))$

$\cos(x)\sec^{2}(\sin (x))$

$\cos(x)\sec(\sin (x))$

$\sec ^{2}(\cos (x))$

Correct answer:

$\cos(x)\sec^{2}(\sin (x))$

Explanation:

Because we are differentiating a function within another function, we must use the chain rule, which states that

Given the equation

$h(x)=\tan(\sin (x))$ ,

we can deduce that

$f(x)=\tan(x)$

and

$g(x)=\sin(x)$ .

By plugging these into the chain rule, we conclude that

$h{}'(x)=\cos(x)\sec^{2}(\sin (x))$ .

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Example Question #42 : Chain Rule And Implicit Differentiation

Find the derivative of the function $x^2\tan(x^3+\sin(x))$

Possible Answers:

$x^2\sec^2(x^2+\sin(x))(2x+\cos(x)) +2x\tan(x^2+\sin(x))$

$x^2\sec^2(x^2+\sin(x)) +2x\tan(x^2+\sin(x))$

$x^2\sec(x^2+\sin(x))(2x+\cos(x))$

None of the other answers

$x\sec(x^2+\sin(x))(2x+\sin(x)) +2\sec(x^2+\sin(x))$

Correct answer:

$x^2\sec^2(x^2+\sin(x))(2x+\cos(x)) +2x\tan(x^2+\sin(x))$

Explanation:

We proceed as follows.

$x^2\tan(x^3+\sin(x))$ . (Start)

$[x^2\tan(x^3+\sin(x))]' = x^2[\tan(x^2+\sin(x))]' +[x^2]'\tan(x^2+\sin(x))$ . (Product rule)

$= x^2[\sec^2(x^2+\sin(x))\times[x^2+\sin(x)]'] +2x\tan(x^2+\sin(x))$ . (The first derivative uses the Chain rule. The 2nd one uses the basic power rule.)

$= x^2[\sec^2(x^2+\sin(x))\times2x+\cos(x)] +2x\tan(x^2+\sin(x))$ .

$= x^2\sec^2(x^2+\sin(x))(2x+\cos(x)) +2x\tan(x^2+\sin(x))$ .

Although some factoring could be done at this point, we will not do so.

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Example Question #131 : Computation Of The Derivative

Find the derivative of the following function:

(3x^2-8)^4 .

Possible Answers:

24x(3x^2-8)

4(6x)^3

24x(6x)^3

4(3x^2-8)^3

24x(3x^2-8)^3

Correct answer:

24x(3x^2-8)^3

Explanation:

For a chain rule derivative, we need to work our way inward from the very outermost function. First, we need to do a power rule for the outer exponent. Then, we multiply that by the derivative of the inside.

$(3x^2-8)^4'=4(3x^2-8)^{4-1}(3x^2-8)'=24x(3x^2-8)^3.$

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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 #34 : Chain Rule And Implicit Differentiation

Example Question #35 : Chain Rule And Implicit Differentiation

Example Question #36 : Chain Rule And Implicit Differentiation

Example Question #37 : Chain Rule And Implicit Differentiation

Example Question #38 : Chain Rule And Implicit Differentiation

Example Question #39 : Chain Rule And Implicit Differentiation

Example Question #40 : Chain Rule And Implicit Differentiation

Example Question #41 : Chain Rule And Implicit Differentiation

Example Question #42 : Chain Rule And Implicit Differentiation

Example Question #131 : Computation Of The Derivative

All AP Calculus AB Resources

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