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AP Calculus AB : Derivatives

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Example Questions

Example Question #1041 : Ap Calculus Ab

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

y^2=4sinx+3x^2

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

Implicit differentiation requires taking the derivative of everything in our equation, including all variables and numbers. Any time we take a derivative of a function with respect to , we need to implicitly write $\frac{dy}{dx}$ after it. Hence, the name of this method. Then, we solve for $\frac{dy}{dx}.$

$(y^2=4sinx+3x^2)'=2y\frac{dy}{dx}=4cosx+6x.$

$\frac{dy}{dx}=\frac{4cosx+6x}{2y}=\frac{2cosx+3x}{y}.$

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Example Question #3 : Implicit Differentiation

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the derivative of the function, we must use implicit differentiation, which is an application of the chain rule. We start by taking the derivative of the function with respect to x, noting that whenever we take a derivative of y, it is with respect to x, so we denote it as .

$y'=\frac{\mathrm{d} }{\mathrm{d} x}(xy^2)+\frac{\mathrm{d} }{\mathrm{d} x}(2\tan{y})$

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

Bringing the terms with to one side and factoring it out, we get

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

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

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Example Question #4 : Implicit Differentiation

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

y=x^3y+y^2

Possible Answers:

$y'=\frac{4x^2y}{1-x^2-2y}$

$y'=\frac{3x^2y}{1-x^2-y}$

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

$y'=\frac{3x^2y}{1-x^3-2y}$

Correct answer:

$y'=\frac{3x^2y}{1-x^3-2y}$

Explanation:

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

y'=y'x^3+3x^2y+2yy'

Bringing the terms with to one side and factoring it out, we get

y'(1-x^3-2y)=3x^2y

$y'=\frac{3x^2y}{1-x^3-2y}$

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Example Question #1 : Implicit Differentiation

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

y+x^2+e^xy^2=2y^2

Possible Answers:

$\frac{-2x-e^xy^2}{1-2ye^x+4y}$

$\frac{-2x-e^xy^2}{2ye^x-4y}$

$\frac{-2x+e^xy^2}{1+2ye^x-4y}$

$\frac{-2x-e^xy^2}{1+2ye^x-4y}$

$\frac{2x-e^xy^2}{1+2ye^x-4y}$

Correct answer:

$\frac{-2x-e^xy^2}{1+2ye^x-4y}$

Explanation:

To find $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we must take the derivative of both sides of the equation with respect to x. When we do this, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}+2x+e^xy^2+2ye^x\frac{\mathrm{d}y }{\mathrm{d} x}=4y\frac{\mathrm{d} y}{\mathrm{d} x}$

The derivatives were found using the following rules:

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

Note that for every derivative of a function of y with respect to x, the chain rule was used, which accounts for $\frac{\mathrm{d} y}{\mathrm{d} x}$ appearing.

Algebraic simplification gets us our final answer,

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{-2x-e^xy^2}{1+2ye^x-4y}$

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Example Question #6 : Implicit Differentiation

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

$\cos(xy)=2x$

Possible Answers:

$-\frac{2+y\sin(xy)}{x\sin(xy)}$

$-\frac{2+x\sin(xy)}{y\sin(xy)}$

$-\frac{2+\sin(xy)-y}{x}$

$\frac{2+x\sin(xy)}{y\sin(xy)}$

$\frac{2+y\sin(xy)}{x\sin(xy)}$

Correct answer:

$-\frac{2+y\sin(xy)}{x\sin(xy)}$

Explanation:

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

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

The following derivative rules were used:

$\frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ , $\frac{\mathrm{d} }{\mathrm{d} x} ax=a$

Note that the chain rule was used for both the cosine function (which contains an inner product of two functions), and for the derivative of y, whose derivative with respect to x we want to solve for.

Solving, we get

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

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Example Question #1 : Implicit Differentiation

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

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

Possible Answers:

$\frac{2ye^{2x}+3x^2}{y^2-e^{2x}}$

$\frac{2ye^{2x}-3x^2}{y^2+e^{2x}}$

$\frac{2ye^{2x}-y^2}{3x^2-e^{2x}}$

$\frac{y^2-e^{2x}}{2ye^{2x}-3x^2}$

$\frac{2ye^{2x}-3x^2}{y^2-e^{2x}}$

Correct answer:

$\frac{2ye^{2x}-3x^2}{y^2-e^{2x}}$

Explanation:

To find the derivative of y with respect to x, we must take the differentiate both sides of the equation with respect to x:

$y^2 \frac{\mathrm{d} y}{\mathrm{d} x}+3x^2=2ye^{2x}+e^{2x}\frac{\mathrm{d} y}{\mathrm{d} x}$

The following rules were used:

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

Note that the chain rule was used everywhere we took the derivative of a function containing y, as well as in the exponential function. The product rule was used because both x and y are both functions of x.

Using algebra to solve, we get

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{2ye^{2x}-3x^2}{y^2-e^{2x}}$

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Example Question #1 : Implicit Differentiation

Find $\frac{\mathrm{d} \alpha}{\mathrm{d} x}$ , where $\alpha$ is a function of x:

$\alpha^2x^2+2\alpha x=\alpha$

Possible Answers:

$\frac{2\alpha^2 x+2 \alpha}{1+2\alpha x^2+2x}$

$\frac{2\alpha^2 x+2 \alpha}{1-2\alpha x^2 -2x}$

$\frac{2\alpha x^2+2 \alpha}{1-2\alpha^2 x -2x}$

None of the other answers

Correct answer:

$\frac{2\alpha^2 x+2 \alpha}{1-2\alpha x^2 -2x}$

Explanation:

To find the derivative of $\alpha$ with respect to x, we must take the differentiate both sides of the equation with respect to x:

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

The following derivative rules were used:

Note that the chain rule was used for the derivative of any function containing $\alpha$ , whose derivative with respect to x we want to solve for.

Solving, we get

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

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Example Question #1 : Implicit Differentiation

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

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To solve this using implicit differentiation, we must always treat y as a function of x, and therefore when we differentiate y with respect to x, we denote it as

Step by step, we get the following:

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

This resulted from the product rule and chain rule

The next steps are:

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

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

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

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

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Example Question #11 : Implicit Differentiation

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

$y=y\cos(x^3)$

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

To find the derivative of the function, we use implicit differentiation, where we always treat y as a function of x, and denoting any derivative of y with respect to x as

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

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

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

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

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Example Question #591 : Derivatives

Find $\frac{\mathrm{d} y}{\mathrm{d} x}$ from the following equation:

$e^{2y}+x^2y^2=y$

Possible Answers:

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

$\frac{-2xy^2-2x^2y}{2e^{2y}-1}$

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

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

Correct answer:

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

Explanation:

To find the derivative of y with respect to x, we must take the derivative with respect to x of both sides of the equation:

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

The derivatives were found using the following rules:

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

Notice the chain rule was used for every function containing y, because y is a function of x and its whose derivative we are interested in isolating.

Using algebra to solve, our final answer is

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

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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 #1041 : Ap Calculus Ab

Example Question #3 : Implicit Differentiation

Example Question #4 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #6 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #1 : Implicit Differentiation

Example Question #11 : Implicit Differentiation

Example Question #591 : Derivatives

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