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AP Calculus AB : Derivatives

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Example Questions

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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Example Question #444 : Ap Calculus Ab

Find the derivative of the following function:

$\sqrt{1-sin^2x}$ .

Possible Answers:

cos(x)

-cos(x)

-sin(x)

sin(x)

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

Correct answer:

-sin(x)

Explanation:

Here, we need to take the derivative of a square root function first. Then, we will multiply that by the derivative of the function under the square root.

$\sqrt{1-sin^2x}'=\frac{1}{2\sqrt{1-sin^2x}}*-2sinx*cosx.$

Remember: -sin^2x'=-2sinx*cosx by another chain rule.

Now, let's make use of our trigonometric identities:

$sin^2x+cos^2x=1. \; 1-sin^2x=cos^2x$

Therefore, we can simplify our answer:

$\frac{-2sinxcosx}{2\sqrt{cos^2x}}=\frac{-sinxcosx}{cosx}=-sinx.$

Note, you could have used this trigonometric identity from the very beginning, making the problem much easier!

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Example Question #42 : Chain Rule And Implicit Differentiation

Find $\frac{dy}{dx}.$

x^2+3y-sinx=2.

Possible Answers:

$\frac{dy}{dx}=\frac{2-cosx-2x}{3}$

$\frac{dy}{dx}=\frac{cosx-2x}{3}$

$\frac{dy}{dx}=\frac{-cosx-2x}{3}$

$\frac{dy}{dx}=\frac{2+cosx-2x}{3}$

$\frac{dy}{dx}=cosx-2x$

Correct answer:

$\frac{dy}{dx}=\frac{cosx-2x}{3}$

Explanation:

For implicit differentiation, we need to take a derivative from left to right of our function. The only difference is that any time we take the derivative of a function, we need to explicitly write $\frac{dy}{dx}$ after it. Then, we will use algebra to solve for that $\frac{dy}{dx}.$

$(x^2+3y-sinx=2)'=2x+3\frac{dy}{dx}-cosx=0.$

$\frac{dy}{dx}=\frac{cosx-2x}{3}.$ Here, just move everything other than the term we want to the right. Then, divide by the coefficient.

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Example Question #43 : Chain Rule And Implicit Differentiation

Find $\frac{dy}{dx}$ for the following function:

x^3-2y^2+3x=4

Possible Answers:

$\frac{dy}{dx}=\frac{3-3x^2}{4}$

$\frac{dy}{dx}=\frac{3-3x^2}{2y^2}$

$\frac{dy}{dx}=\frac{3+3x^2}{4y}$

$\frac{dy}{dx}=\frac{3+3x^2}{2y^2}$

$\frac{dy}{dx}=\frac{3+3x^2}{4}$

Correct answer:

$\frac{dy}{dx}=\frac{3+3x^2}{4y}$

Explanation:

For implicit differentiation, we need to take the derivative of every term, including numbers. Whenever we take a derivative of a function of , we need to multiply it by $\frac{dy}{dx}$ . Then, we will rearrange the equation to solve for $\frac{dy}{dx}.$

$(x^3-2y^2+3x=4)'=3x^2-4y\frac{dy}{dx}+3=0$

$\frac{dy}{dx}=\frac{-3x^2-3}{-4y}=\frac{3x^2+3}{4y}.$

In the last step, I canceled all of the negative signs.

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Example Question #141 : Computation Of The Derivative

Determine the derivative of $f(x)=2\tan ^2(x^2)$

Possible Answers:

$4x\tan(x^2)$

$8x\tan(x^2)\sec^2(x^2)$

$8x\sec^2(x^2)$

$4\tan(x^2)\sec^2(x^2)$

$4x\sec^2(x^2)$

Correct answer:

$8x\tan(x^2)\sec^2(x^2)$

Explanation:

This is a pure problem on understanding how chain rules work for derivatives.

First thing we need to remember is that the derivative of $\tan(x)$ is $\sec^2(x)$ .

When we are taking the derivative of $f(x)=2\tan ^2(x^2)$ , we can first pull out the 2 in the front and we treat $\tan^2(x^2)$ as $[\tan(x^2)]^2$ .

This way, the derivative will become $2*2*\tan(x^2)*\frac{\mathrm{d} tan(x^2)}{\mathrm{d} x}$ ,

which is $4\tan(x^2)*(2x\sec(x^2))$ .

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AP Calculus AB : Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

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

Example Question #444 : Ap Calculus Ab

Example Question #42 : Chain Rule And Implicit Differentiation

Example Question #43 : Chain Rule And Implicit Differentiation

Example Question #141 : Computation Of The Derivative

All AP Calculus AB Resources

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