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AP Calculus AB : Applications of Derivatives

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Example Questions

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Example Question #28 : Implicit Differentiation

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ :

x^2+xy=4

Possible Answers:

-2x-y

$\frac{2x-y}{x}$

$\frac{1}{x}$

$\frac{-2x-y}{x}$

Correct answer:

$\frac{-2x-y}{x}$

Explanation:

To find $\frac{\mathrm{d} y}{\mathrm{d} x}$ we must use implicit differentiation, which is an application of the chain rule.
Taking $\frac{\mathrm{d} }{\mathrm{d} x}$ of both sides of the equation, we get

$2x+y+x\frac{\mathrm{d} y}{\mathrm{d} x}=0$

using the following rules:

Note that for every derivative of a function with y, the additional term $\frac{\mathrm{d} y}{\mathrm{d} x}$ appears; this is because of the chain rule, where y=g(x), so to speak, for the function it appears in.

Using algebra to solve for $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we get

$\frac{-2x-y}{x}$ .

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Example Question #29 : Implicit Differentiation

Use implicit differentiation to calculate the equation of the line tangent to the equation $2x^{2}+y^{2}=9$ at the point (2,1).

Possible Answers:

y=-4x+9

y=-4x-9

y=4x+9

y=4x-9

Correct answer:

y=-4x+9

Explanation:

Differentiate both sides of the equation: $\frac{\mathrm{d} }{\mathrm{d} x}(2x^{2}+y^{2}=9)$

Simplify: $4x+\frac{\mathrm{d} }{\mathrm{d} x}y^{2}=0$

Use implicit differentiation to differentiate the y term: $4x+2y\frac{dy}{dx}=0$

Subtract 4x from both sides of the equation: $2y\frac{dy}{dx}=-4x$

Divide both sides of the equation by 2y: $\frac{dy}{dx}=\frac{-4x}{2y}=\frac{-2x}{y}$

Plug in the appropriate values for x and y to find the slope of the tangent line: $m=\frac{-2(2)}{(1)}=-4$

Use slope-intercept form to solve for the equation of the tangent line: y=mx+b=y=-4x+b

Plug in the appropriate values of x and y into the equation, to find the equation of the tangent line: 1=(2)(-4)+b

Solve for b: 1=-8+b

b=9

Solution: y=-4x+9

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Example Question #81 : Applications Of Derivatives

Find $\frac{\mathrm{d} z}{\mathrm{d} x}$ , where is a function of x.

Possible Answers:

$\frac{3-2z}{x}$

$\frac{3-z}{x+2z}$

$\frac{-z}{x+2z}$

Correct answer:

$\frac{3-z}{x+2z}$

Explanation:

To find $\frac{\mathrm{d} z}{\mathrm{d} x}$ we must use implicit differentiation, which is an application of the chain rule.
Taking $\frac{\mathrm{d} }{\mathrm{d} x}$ of both sides of the equation, we get

$z+x\frac{\mathrm{d} z}{\mathrm{d} x}+2z\frac{\mathrm{d} z}{\mathrm{d} x}=3$

and the derivatives were found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$
Note that for every derivative of a function with z, the additional term $\frac{\mathrm{d} z}{\mathrm{d} x}$ appears; this is because of the chain rule, where z=g(x), so to speak, for the function it appears in.

Using algebra to solve, we get

$\frac{\mathrm{d} z}{\mathrm{d} x}=\frac{3-z}{x+2z}$

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Example Question #31 : Implicit Differentiation

Find the normal line of the curve $\frac{x^{2}}{2}+y^{2}=4$ at the point $(1,\frac{\sqrt{14}}{2})$

Possible Answers:

$y=x\sqrt{14}+\frac{\sqrt{14}}{2}$

$y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

$y=x\sqrt{14}-{\sqrt{14}}$

$y=2x\sqrt{14}-\frac{\sqrt{14}}{2}$

Correct answer:

$y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

Explanation:

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

Use implicit differentiation to calculate the slope of the tangent line:

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

$\frac{2x}{2}+2y\frac{dy}{dx}=0$

Simplify: $x+2y\frac{dy}{dx}=0$

Subtract x from both sides of the equation: $2y\frac{dy}{dx}=-x$

Divide both sides of the equation by 2y: $\frac{dy}{dx}=\frac{-x}{2y}$

Plug in the x and y values from the point $(1,\frac{\sqrt{14}}{2})$ into the equation, to calculate the slope of the tangent line: $\frac{dy}{dx}=\frac{-1}{\frac{2\sqrt{14}}{2}}=\frac{-1}{\sqrt{14}}=\frac{-\sqrt{14}}{14}$

Take the negative reciprocal of the slope of the tangent line to calculate the slope of the normal line: $m=-\frac{1}{\frac{-\sqrt{14}}{14}}=\sqrt{14}$

Plug the slope of the normal line into point slope form: $y-\frac{\sqrt{14}}{2}=(x-1)\sqrt{14}$

Add $\frac{\sqrt{14}}{2}$ to both sides to isolate y: $y=(x-1)\sqrt{14}+\frac{\sqrt{14}}{2}$

Simplify the equation: $y=x\sqrt{14}-\sqrt{14}+\frac{\sqrt{14}}{2}$

Simplify further: $y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

Solution: $y=x\sqrt{14}-\frac{\sqrt{14}}{2}$

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Example Question #21 : Calculus I — Derivatives

Find $f^{'}(2)$ if the function f(x) is given by

f(x) = ln(x)

Possible Answers:

$\frac{3}{2}$

$\frac{1}{2}$

Correct answer:

$\frac{1}{2}$

Explanation:

To find the derivative at x=2 , we first take the derivative of f(x) . By the derivative rule for logarithms,

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

Plugging in x=2 , we get

$f'(2) = \frac{1}{2}$

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Example Question #22 : Calculus I — Derivatives

Find the derivative of the following function at the point x=3 .

f(x) = 4x^3+x+3

Possible Answers:

106

108

109

110

107

Correct answer:

109

Explanation:

Use the power rule on each term of the polynomial to get the derivative,

f'(x) = (3)(4)x^2+1

Now we plug in x=3

f'(3) = (12)(3)^2 + 1 = 109

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Example Question #23 : Calculus I — Derivatives

Let $f(x)=x^2\sin(x^2)$ . What is $f'\left (\frac{\sqrt{\pi}}{2} \right )$ ?

Possible Answers:

$\frac{\sqrt{2\pi}(\pi + 4)}{8}$

$\frac{{2\pi}(\sqrt{\pi} + 2)}{4}$

$\frac{\sqrt{2\pi}(\pi + \sqrt{2})}{4}$

$\frac{(\pi + \sqrt{2})}{8}$

$\frac{\sqrt{2\pi}(\pi + 2)}{4}$

Correct answer:

$\frac{\sqrt{2\pi}(\pi + 4)}{8}$

Explanation:

We need to find the first derivative of f(x). This will require us to apply both the Product and Chain Rules. When we apply the Product Rule, we obtain:

$f'(x)=\sin(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]+x^2\cdot\frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]$

In order to find the derivative of $\sin(x^2)$ , we will need to employ the Chain Rule.

$\frac{\mathrm{d} }{\mathrm{d} x}[\sin(x^2)]=\cos(x^2)\cdot\frac{\mathrm{d} }{\mathrm{d} x}[x^2]=\cos(x^2)\cdot2x$

$f'(x)=\sin(x^2)\cdot2x + x^2\cdot\cos(x^2)\cdot2x$

We can factor out a 2x to make this a little nicer to look at.

$f'(x)=2x(\sin(x^2)+x^2\cos(x^2))$

Now we must evaluate the derivative when x = $\frac{\sqrt{\pi}}{2}$ .

$f'\left (\frac{\sqrt{\pi}}{2} \right )=2\cdot\frac{\sqrt{\pi}}{2}(\sin\frac{\pi}{4}+\frac{\pi}{4}\cos\frac{\pi}{4})$

$=\sqrt{\pi}(\frac{\sqrt{2}}{2}+\frac{\pi\sqrt{2}}{8})=\sqrt{2\pi}(\frac{1}{2}+\frac{\pi}{8})=\frac{\sqrt{2\pi}(\pi + 4)}{8}$

The answer is $\frac{\sqrt{2\pi}(\pi + 4)}{8}$ .

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AP Calculus AB : Applications of Derivatives

Study concepts, example questions & explanations for AP Calculus AB

All AP Calculus AB Resources

Example Questions

Example Question #28 : Implicit Differentiation

Example Question #29 : Implicit Differentiation

Example Question #81 : Applications Of Derivatives

Example Question #31 : Implicit Differentiation

Example Question #21 : Calculus I — Derivatives

Example Question #22 : Calculus I — Derivatives

Example Question #23 : Calculus I — Derivatives

All AP Calculus AB Resources

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