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AP Calculus AB : Computation of the Derivative

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Example Questions

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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Example Question #441 : Ap Calculus Ab

Find the derivative of the following function:

sin(tanx) .

Possible Answers:

cos(sec^2x)

-cos(tanx)

sec^2xcos(tanx)

cos(tanx)

sin(sec^2x)

Correct answer:

sec^2xcos(tanx)

Explanation:

For a chain rule derivative, we take the derivative of the outside function (leaving the inside function unchanged). Then, you multiply that by the derivative of the inside.

sin(tanx)'=sin'(tanx)*tanx'=cos(tanx)*sec^2x

=sec^2xcos(tanx).

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Example Question #442 : Ap Calculus Ab

Find the derivative of the following function:

f(x)=ln(3x^7) .

Possible Answers:

21x^6ln(3x^7)

$\frac{3}{x^7}$

$\frac{3}{x}$

$\frac{7}{x}$

$\frac{1}{3x^7}$

Correct answer:

$\frac{7}{x}$

Explanation:

For a chain rule derivative, we need to take the derivative of lnx first. Then, we multiply by the derivative of the inside function. In other words, the general chain derivative of the function is:

$ln(u)'=\frac{u'}{u}.$

$ln(3x^7)=\frac{21x^6}{3x^7}=\frac{7}{x}.$

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Example Question #443 : Ap Calculus Ab

Find the derivative of the following function:

$e^{cotx}$ .

Possible Answers:

$-sec^2xe^{cotx}$

$csc^2xe^{cotx}$

$e^{-csc^2x}$

$-csc^2xe^{cotx}$

$sec^2xe^{cotx}$

Correct answer:

$-csc^2xe^{cotx}$

Explanation:

The exponential function is the only derivative that always returns the original function. Therefore, we only need to multiply that by the derivative of the new exponent.

e^u'=e^u*u' .

$e^{cotx}'=e^{cotx}*-csc^2x=-csc^2xe^{cotx}.$

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AP Calculus AB : Computation of the Derivative

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

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

Example Question #441 : Ap Calculus Ab

Example Question #442 : Ap Calculus Ab

Example Question #443 : Ap Calculus Ab

All AP Calculus AB Resources

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