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AP Calculus AB : 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 #11 : Implicit Differentiation

Given that y=y(x) , compute the derivative of the following function:

$y=\tan(xy^2)$

Possible Answers:

$y'=\frac{y^2\sec^2(xy^2)}{1-2xy\sec^2(y^2)}$

$y'=\frac{y^2\sec^2(xy^2)}{1-2xy\sec^2(xy^2)}$

$y'=\frac{y^2\sec^2(xy^2)}{1-2xy\tan^2(xy^2)}$

$y'=\frac{y^2\sec^2(xy^2)}{1+2xy\sec^2(xy^2)}$

Correct answer:

$y'=\frac{y^2\sec^2(xy^2)}{1-2xy\sec^2(xy^2)}$

Explanation:

To find the derivative of the function, we use implicit differentiation, which is an application of the chain rule. We use this because y=y(x) , and any derivative with respect to is $\frac{\mathrm{d} y}{\mathrm{d} x}$ (or ).

First, we use the chain rule combined with the product rule in taking the derivative of y

$y'=\sec^2(xy^2)(y^2+2xy(y'))$

Then, we expand in order to isolate the terms with

$y'=y^2\sec^2(xy^2)+2xyy'\sec^2(xy)$

$y'-2xyy'\sec^2(xy^2)=y^2\sec^2(xy^2)$

Finally, we factor out a

$y'(1-2xy\sec^2(xy^2))=y^2\sec^2(xy^2)$

$y'=\frac{y^2\sec^2(xy^2)}{1-2xy\sec^2(xy^2)}$

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Example Question #12 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function using implicit differentiation

$y=\sin{xy}-5y^4+\tan{x}$

Possible Answers:

$y'=\frac{y\cos{xy}+\sec^2{x}}{1-x\sin{xy}+20y^3}$

$y'=\frac{y\cos{xy}+\sec{x}}{1-x\cos{xy}+20y^3}$

$y'=\frac{y\cos{xy}+\sec^2{x}}{1+x\cos{xy}+10y^3}$

$y'=\frac{y\cos{xy}+\sec^2{x}}{1-x\cos{xy}+20y^3}$

Correct answer:

$y'=\frac{y\cos{xy}+\sec^2{x}}{1-x\cos{xy}+20y^3}$

Explanation:

To find the derivative with respect to y, we must use implicit differentiation, which is an application of the chain rule.

$y'=\cos(xy)(y+xy')-20y^3y'+\sec^2(x)$

$y'=y\cos(xy)+xy'\cos(xy)-20y^3y'+\sec^2(x)$

$y'-xy'\cos(xy)+20y^3y'=y\cos(xy)+\sec^2(x)$

$y'(1-x\cos(xy)+20y^3)=y\cos(xy)+\sec^2(x)$

$y'=\frac{y\cos{xy}+\sec^2{x}}{1-x\cos{xy}+20y^3}$

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Example Question #71 : Applications Of Derivatives

Differentiate the following implicit function:

(x-2)^2+y^2=121

Possible Answers:

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

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

$\frac{dy}{dx}=\frac{2x-1}{y}$

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

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

Correct answer:

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

Explanation:

For this problem we are asked to find $\frac{dy}{dx}$ , or the rate of change in y with respect to x.

To do this we take the derivative of each variable and to differentiate between the two, we will write dx or dy after.

(x-2)^2+y^2=121

would then become

(2x-4)dx+(2y)dy=0

We note that the derivative of a constant is still zero.

We must now rewrite this function in the form $\frac{dy}{dx}$

(2y)dy=-(2x-4)dx

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

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

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Example Question #601 : Derivatives

Find the implicit derivative, $\frac{dy}{dx}$ a circle centered at (2,-1) with radius r=2 .

Possible Answers:

$\frac{dy}{dx}=\frac{x+2}{y-1}$

$\frac{dy}{dx}=2x+2y-1$

$\frac{dy}{dx}=-\frac{x+2}{y-1}$

$\frac{dy}{dx}=x+y-1$

$\frac{dy}{dx}=-\frac{x-2}{y+1}$

Correct answer:

$\frac{dy}{dx}=-\frac{x-2}{y+1}$

Explanation:

The equation of a circle centered at (2,-1) with radius r=2 is (x-2)^2+(y+1)^2=4 .

We first expand our equation to simplify the derivative.

x^2-4x+4+y^2+2y+1=0

x^2-4x+y^2+2y=-5

Take the derivatives of x and y we get:

(2x-4)dx+(2y+2)dy=0

Since the derivative of a constant is zero.

Next we must rewrite our equation in terms of $\frac{dy}{dx}$ :

(2y+2)dy=-(2x-4)dx

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

Simplifying:

$\frac{dy}{dx}=-\frac{x-2}{y+1}$

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Example Question #21 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function

$y=\sin(xy)+y^3$

Possible Answers:

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-3y^3}$

$y'=\frac{y\cos(xy)}{1-x\sin(xy)-3y^2}$

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-3y^2}$

$y'=\frac{x\cos(xy)}{1-x\cos(xy)-3y^2}$

Correct answer:

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-3y^2}$

Explanation:

To find the derivative with respect to y, we use implicit differentiation, which is an application of the chain rule.

$y=\sin(xy)+y^3$

$y'=\cos(xy)(y+xy')+3y^2y'$

$y'-xy'\cos(xy)-3y^2y'=y\cos(xy)$

$y'(1-x\cos(xy)-3y^2)=y\cos(xy)$

$y'=\frac{y\cos(xy)}{1-x\cos(xy)-3y^2}$

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Example Question #22 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function

$y=\tan(y^2)+xy^5$

Possible Answers:

$y'=\frac{5y^4}{1-2y\sec^2(y^2)-5xy^4}$

$y'=\frac{y^5}{1-2y\sec^2(y^2)-5xy^4}$

$y'=\frac{y^5}{1+2y\sec^2(y^2)+5xy^4}$

$y'=\frac{y^5}{1-2y\tan^2(y^2)-5xy^4}$

Correct answer:

$y'=\frac{y^5}{1-2y\sec^2(y^2)-5xy^4}$

Explanation:

To find the derivative with respect to y, we use implicit differentiation, which is an application of the chain rule

$y=\tan(y^2)+xy^5$

$y'=\sec^2(y^2)(2yy')+y^5+5xy^4(y')$

$y'-2yy'\sec^2(y^2)-5xy^4y'=y^5$

$y'(1-2y\sec^2(y^2)-5xy^4)=y^5$

$y'=\frac{y^5}{1-2y\sec^2(y^2)-5xy^4}$

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Example Question #23 : Implicit Differentiation

Given that y=y(x) , find the derivative of the function

$y=2xy^3+x\cos(y)$

Possible Answers:

$y'=\frac{2y^3+\cos(y)}{1-x\sin(y)-6xy^2}$

$y'=\frac{2y^3+\sin(y)}{1+x\sin(y)-6xy^2}$

$y'=\frac{2y^3+\cos(y)}{1+x\sin(y)-6xy^2}$

$y'=\frac{2y^3+\cos(y)}{1+\sin(xy)-6xy^2}$

Correct answer:

$y'=\frac{2y^3+\cos(y)}{1+x\sin(y)-6xy^2}$

Explanation:

To find the derivative with respect to y, we use implicit differentiation, which is an application of the chain rule

$y=2xy^3+x\cos(y)$

$y'=2y^3+6xy^2y'+\cos(y)-x\sin(y)y'$

$y'+xy'\sin(y)-6xy^2y'=2y^3+\cos(y)$

$y'(1+x\sin(y)-6xy^2)=2y^3+\cos(y)$

$y'=\frac{2y^3+\cos(y)}{1+x\sin(y)-6xy^2}$

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Example Question #21 : Implicit Differentiation

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

$xy+\ln(y)=x$

Possible Answers:

$\frac{y-y^2}{xy+1}$

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

$\frac{y-y^2}{xy+y}$

$\frac{-y}{xy+1}$

Correct answer:

$\frac{y-y^2}{xy+1}$

Explanation:

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

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

The derivative was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} \ln(x)=\frac{1}{x}$

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

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

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Example Question #27 : Implicit Differentiation

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

$3y+x\tan(y)=2$

Possible Answers:

$\frac{\tan(y)}{3+x\sec^2(y)}$

$\frac{-\tan(y)}{3-x\sec^2(y)}$

$\frac{-\tan(y)}{3+x\sec^2(y)}$

$\frac{-\tan(y)}{3+\sec^2(y)}$

Correct answer:

$\frac{-\tan(y)}{3+x\sec^2(y)}$

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

$3 \frac{\mathrm{d} y}{\mathrm{d} x} +\tan(y)+x\sec^2(y)\frac{\mathrm{d} y}{\mathrm{d} x}=0$

The following derivative rules were used:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)$

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

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-\tan(y)}{3+x\sec^2(y)}$

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Example Question #81 : Applications Of Derivatives

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

x^2+xy=4

Possible Answers:

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

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

-2x-y

$\frac{1}{x}$

Correct answer:

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

Explanation:

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

using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

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

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

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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 #11 : Implicit Differentiation

Example Question #12 : Implicit Differentiation

Example Question #71 : Applications Of Derivatives

Example Question #601 : Derivatives

Example Question #21 : Implicit Differentiation

Example Question #22 : Implicit Differentiation

Example Question #23 : Implicit Differentiation

Example Question #21 : Implicit Differentiation

Example Question #27 : Implicit Differentiation

Example Question #81 : Applications Of Derivatives

All AP Calculus AB Resources

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