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AP Calculus AB : Computation of the Derivative

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Example Questions

Example Question #61 : Chain Rule And Implicit Differentiation

Find the derivative of the function

$w(x)=\sin(\tan{x})$

Possible Answers:

$w'(x)=\sec^2(x)-\cos(\tan{x})$

$w'(x)=\sec^2(x)\cos(\tan{x})$

$w'(x)=\sec^2(x)\sin(\tan{x})$

$w'(x)=\sec(x)\cos(\tan{x})$

Correct answer:

$w'(x)=\sec^2(x)\cos(\tan{x})$

Explanation:

To find the derivative of the function, you must apply the chain rule, which is as follows:

$\frac{\mathrm{d} }{\mathrm{d} x}(p(g(x)))=p'(g(x))(g'(x))$

Using the function from the problem statement, we have that

$g(x)=\tan(x)$ and $p(g(x))=\sin(\tan(x))$

Following the rule, we get

$w'(x)=\sec^2(x)\cos(\tan(x))$

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Example Question #62 : Chain Rule And Implicit Differentiation

$f(x) = \cos (x+ \ln x)$

Find f'(x) .

Possible Answers:

None of the other choices gives the correct response.

$f'(x)= \frac{(x+1)\sin \left ( x + \ln x \right )}{x}$

$f'(x)= \frac{(-x- 1)\sin \left ( x + \ln x \right )}{x}$

$f'(x)= \frac{(-x- 1)\sin \left ( x - \ln x \right )}{x}$

$f'(x)= \frac{(x+1)\sin \left ( x - \ln x \right )}{x}$

Correct answer:

$f'(x)= \frac{(-x- 1)\sin \left ( x + \ln x \right )}{x}$

Explanation:

Let $u(x) = x + \ln x$

Then

$f(x) = \cos u$

and

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x} \cos (u(x))$

Apply the chain rule:

$f'(x)= \frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} u} \cos u$

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}( x + \ln x) \cdot( - \sin u)$

Substitute back for :

$f'(x)= \frac{\mathrm{d} }{\mathrm{d} x}( x + \ln x) \cdot( - \sin \left ( x + \ln x \right ))$

Apply the sum rule:

$f'(x)=\left ( \frac{\mathrm{d} }{\mathrm{d} x}( x ) + \frac{\mathrm{d} }{\mathrm{d} x}( \ln x) \right )\cdot( - \sin \left ( x + \ln x \right ))$

$f'(x)=\left ( 1 + \frac{1}{x} \right )\cdot( - \sin \left ( x + \ln x \right ))$

After some simple algebra:

$f'(x)=\left ( \frac{x+ 1}{x} \right )\cdot( - \sin \left ( x + \ln x \right ))$

$f'(x)= \frac{(-x- 1)\sin \left ( x + \ln x \right )}{x}$

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Example Question #161 : Computation Of The Derivative

is a function of . Solve for in this differential equation:

$\frac{\mathrm{d} y}{\mathrm{d} x}= \frac{xy}{x^{2}+ 2}$

Possible Answers:

$y =\frac{C\sqrt{ x^{2}+ 2} }{ x^{2}+ 2}$

$y =C \sqrt{ x^{2}+ 2}$

$y =C e^{\sqrt{ \frac{1}{3}x^{3}+ 2x}}$

$y = \frac{1}{3}x^{3}+ 2x+ C$

$y =C e^{\sqrt{ x^{2}+ 2}}$

Correct answer:

$y =C \sqrt{ x^{2}+ 2}$

Explanation:

The expressions with can be separated from those with by multiplying both sides by $\frac{1}{y}\cdot \mathrm{d} x$ :

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{1}{y}\cdot \mathrm{d} x = \frac{xy}{x^{2}+ 2} \cdot \frac{1}{y}\cdot \mathrm{d} x$

$\frac{\mathrm{d} y}{y} = \frac{x \mathrm{\; d} x}{x^{2}+ 2}$

Find the indefinite integral of both sides:

$\int \frac{ \mathrm{d} y }{y}=\int \frac{x \mathrm{ \; d} x}{x^{2}+ 2 }$

$\ln y =\int \frac{x \mathrm{\; d} x}{x^{2}+ 2 }$

Set $u = x^{2}+ 2$ . Then $\mathrm{d} u = 2x \; \mathrm{ d} x$ , or $x \; \mathrm{ d} x =\frac{1}{2} \mathrm{d} u$ , and

$\ln y =\frac{1}{2}\int \frac{ { d} u}{u }$

$\ln y =\frac{1}{2} ( \ln u + C)$

$\ln y =\frac{1}{2} \ln u + C_{1}$

$\ln y = \ln \sqrt{u} + C_{1}$

Substitute back:

$\ln y = \ln \sqrt{ x^{2}+ 2} + C_{1}$ ;

Raise to both powers:

$\ln y = \ln \sqrt{ x^{2}+ 2} + C_{1}$

$e^{ \ln y }=e^{ \ln \sqrt{ x^{2}+ 2} + C_{1}}$

$y =C_{2}e^{ \ln \sqrt{ x^{2}+ 2} }$

$y =C_{2} \sqrt{ x^{2}+ 2}$ .

The correct choice is $y =C \sqrt{ x^{2}+ 2}$

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Example Question #162 : Computation Of The Derivative

Find the derivative using the chain rule.

$\cos (3x^2)$

Possible Answers:

$-6x\sin (3x^2)$

$-6x\cos (3x^2)$

$6x\sin (3x^2)$

$6x\cos(3x^2)$

Correct answer:

$-6x\sin (3x^2)$

Explanation:

Use the chain rule to find the derivative: $\frac{d}{dx}f(g(x))=g'(x)f'(g(x))$

Thus, $\frac{d}{dx}\cos (3x^2)=-6x\sin (3x^2)$

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Example Question #161 : Computation Of The Derivative

Find the derivative using the chain rule.

$\tan (5x)$

Possible Answers:

$5\sec (5x)$

$\sec^2(5x)$

$5\sec^2(5x)$

$\sec^2(5)$

Correct answer:

$5\sec^2(5x)$

Explanation:

Use the chain rule to find the derivative.

$\frac{d}{dx}\tan (5x)=5\sec^2(5x)$

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Example Question #162 : Computation Of The Derivative

Find the derivative using the chain rule.

$\sin (-3x)$

Possible Answers:

$3\cos (3x)$

$-3\sin (-3x)$

$-3\cos (-3x)$

$3\sin (3x)$

Correct answer:

$-3\cos (-3x)$

Explanation:

Use the chain rule to find the derivative.

$\frac{d}{dx}\sin (-3x)=-3\cos (-3x)$

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Example Question #163 : Computation Of The Derivative

Find the derivative using the chain rule.

$\cos (-2x)$

Possible Answers:

$-2\cos (-2x)$

$2\cos (-2x)$

$-2\sin (-2x)$

$2\sin (-2x)$

Correct answer:

$2\sin (-2x)$

Explanation:

Use the chain rule to find the derivative.

$\frac{d}{dx}\cos (-2x)=2\sin(-2x)$

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Example Question #254 : Derivatives

$y+ \cos y = x^{2}$ .

Which of the following expressions is equal to $\frac{\mathrm{d} y}{\mathrm{d} x}$ ?

Possible Answers:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 + \tan y }$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{ \sin y -1 }$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \tan y }$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \sin y }$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \cos y }$

Correct answer:

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \sin y }$

Explanation:

Differentiate both sides with respect to :

$y+ \cos y = x^{2}$

$\frac{\mathrm{d} }{\mathrm{d} x}(y+ \cos y )= \frac{\mathrm{d} }{\mathrm{d} x}(x^{2})$

$\frac{\mathrm{d} }{\mathrm{d} x}(y+ \cos y )=2x$

By the sum rule:

$\frac{\mathrm{d} }{\mathrm{d} x}(y )+ \frac{\mathrm{d} }{\mathrm{d} x}( \cos y ) =2x$

$\frac{\mathrm{d}y }{\mathrm{d} x} + \frac{\mathrm{d} }{\mathrm{d} x}( \cos y ) =2x$

By the chain rule:

$\frac{\mathrm{d}y }{\mathrm{d} x} + \frac{\mathrm{d} y}{\mathrm{d} x} \frac{\mathrm{d} }{\mathrm{d} y}( \cos y ) =2x$

$\frac{\mathrm{d}y }{\mathrm{d} x} + \frac{\mathrm{d} y}{\mathrm{d} x} \cdot (-\sin y) =2x$

Applying some algebra:

$\frac{\mathrm{d}y }{\mathrm{d} x} \left (1- \sin y \right ) =2x$

$\frac{\frac{\mathrm{d}y }{\mathrm{d} x} \left (1- \sin y \right )}{1 - \sin y } =\frac{2x}{1 - \sin y }$

$\frac{\mathrm{d}y }{\mathrm{d} x}=\frac{2x}{1 - \sin y }$

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Example Question #74 : Chain Rule And Implicit Differentiation

$y^{2}- 4y + 8 = x$

Which of the following is equal to $\frac{\mathrm{d} y}{\mathrm{d} x}$ ?

Possible Answers:

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{y}{ 2y - 4 }$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{x}{ y - 4 }$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ 2y - 4 }$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{x}{ 2y - 4 }$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ y - 4 }$

Correct answer:

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ 2y - 4 }$

Explanation:

Differentiate both sides with respect to :

$y^{2}- 4y + 8 = x$

$\frac{\mathrm{d} }{\mathrm{d} x} (y^{2}- 4y + 8 )=\frac{\mathrm{d} }{\mathrm{d} x} (x)$

Apply the sum, difference, and constant multiple rules:

$\frac{\mathrm{d} }{\mathrm{d} x} (y^{2})- \frac{\mathrm{d} }{\mathrm{d} x} (4y )+\frac{\mathrm{d} }{\mathrm{d} x} ( 8 )=\frac{\mathrm{d} }{\mathrm{d} x} (x)$

In the first term, apply the chain rule; in the second, apply the constant multiple rule:

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d} y} (y^{2})- 4 \frac{\mathrm{d} y}{\mathrm{d} x} +0=1$

Apply the power rule:

$\frac{\mathrm{d} y}{\mathrm{d} x} \cdot 2y - 4 \frac{\mathrm{d} y}{\mathrm{d} x} =1$

Now apply some algebra:

$( 2y - 4 )\frac{\mathrm{d} y}{\mathrm{d} x} =1$

$\frac{\mathrm{d} y}{\mathrm{d} x} =\frac{1}{ 2y - 4 }$

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Example Question #165 : Computation Of The Derivative

We have three functions, g(x)=cos(x), h(x)=sin(x), k(x)=x^2

Find the derivative of f(x)

Given that f(x)=g(h(k(x)))

Possible Answers:

-sin(sin(x^2))*cos(2x)

sin(sin(x^2))*cos(x^2)

-sin(sin(x^2))*cos(x^2)*2x

-sin(cos(2x))

sin(sin(x^2))*cos(x^2)*2x

Correct answer:

-sin(sin(x^2))*cos(x^2)*2x

Explanation:

So now this is a three layer chain rule differentiation. The more functions combine to form the composite function the harder it will be to keep track of the derivative. I find it helpful to lay out each equation and each derivative, so:

g(x)=cos(x), g'(x)=-sin(x)

h(x)=sin(x), h'(x)=cos(x)

k(x)=x^2, k'(x)=2x

Then a three layer chain rule is just the same as a two layer, except... there's one more layer!

f'(x)=g'(h(k(x)))*h'(k(x))*k'(x)

It is still the outermost layer evaluated at the inner layers, and then move another layer in and repeat

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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 #61 : Chain Rule And Implicit Differentiation

Example Question #62 : Chain Rule And Implicit Differentiation

Example Question #161 : Computation Of The Derivative

Example Question #162 : Computation Of The Derivative

Example Question #161 : Computation Of The Derivative

Example Question #162 : Computation Of The Derivative

Example Question #163 : Computation Of The Derivative

Example Question #254 : Derivatives

Example Question #74 : Chain Rule And Implicit Differentiation

Example Question #165 : Computation Of The Derivative

All AP Calculus AB Resources

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