Implicit differentiation - AP Calculus AB

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Question

Let $f(x)=x^2-\frac{1}{1-x^2}$ . Which of the following gives the equation of the line normal to $f(x)$ when ?

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Answer

We are asked to find the normal line. This means we need to find the line that is perpendicular to the tangent line at . In order to find the tangent line, we will need to evaluate the derivative of at .

$f(x)=x^2-\frac{1}{1-x^2}=x^2-(1-x^2)^{-1}$

$f'(x)=2x-(-1)(1-x^2)^{-2}(-2x)$

$f'(x)=2x-2x(1-x^2)^{-2}$

$f'(2)=2(2)-2(2)(1-2^2)^{-2}$

$f'(2)=4-4(\frac{1}{9})=\frac{32}{9}$

The slope of the tangent line at is $\frac{32}{9}$ . Because the tangent line and the normal line are perpendicular, the product of their slopes must equal .

(slope of tangent)(slope of normal) =

$\left ( \frac{32}{9} \right )\cdot \left ( slope\ of\ normal \right )=-1$

$slope\ of\ normal=-\frac{9}{32}$

We now have the slope of the normal line. Once we find a point through which it passes, we will have enough information to derive its equation.

Since the normal line passes through the function at , it will pass through the point $\left ( 2,f(2) \right )$ . Be careful to use the original equation for , not its derivative.

$f(2)=2^2-(1-4)^{-1}=4-(-\frac{1}{3})=\frac{13}{3}$

The normal line has a slope of $-\frac{9}{32}$ and passes through the piont $\left ( 2,\frac{13}{3} \right )$ . We can now use point-slope form to find the equation of the normal line.

$y-\frac{13}{3}=-\frac{9}{32}(x-2)$

Multiply both sides by .

$96y-416=-27(x-2)$

$27x + 96y = 470$

The answer is $27x + 96y = 470$ .

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Question

Use implicit differentiation to find $\frac{\mathrm{d}y }{\mathrm{d} x}$ given

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Answer

We simply differentiate both sides of the equation

(Don't forget the chain rule)

Now we solve for $\frac{\mathrm{d}y }{\mathrm{d} x}$

$xcos(x^2)=\frac{ysin(y^2)dy}{dx}$

$\frac{xcos(x^2)}{ysin(y^2)}=\frac{\mathrm{d}y }{\mathrm{d} x}$

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Question

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

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Answer

Implicit differentiation is very similar to normal differentiation, but every time we take a derivative with respect t , we need to multiply the result by $\frac{dy}{dx}.$ We also differentiate the entire equation from left to right, including any numbers. Then, we solve for that $\frac{dy}{dx}$ for our final answer.

$2x+6y\frac{dy}{dx}=0$

$\frac{dy}{dx}=\frac{-2x}{6y}=\frac{-x}{3y}.$

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Question

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

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Answer

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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Question

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

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Answer

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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Question

Given that , find the derivative of the function with respect to x

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

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Answer

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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Question

Given that , find the derivative of the function

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Answer

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 .

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

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

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

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Question

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

$\cos(xy)=2x$

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Answer

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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Question

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

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

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Answer

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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Question

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

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

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Answer

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:

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

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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Question

Given that , find the derivative of the function

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

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Answer

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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Question

Given that , find the derivative of the following function:

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

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Answer

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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Question

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

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

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Answer

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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Question

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

$y^4=\sin(2x)$

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Answer

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

$4y^3 \frac{\mathrm{d} y}{\mathrm{d} x}=2\cos(2x)$

The derivatives were found using the following rules:

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

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.

Solving for $\frac{\mathrm{d} y}{\mathrm{d} x}$ , we get

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

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Question

Evaluate $\frac{\mathrm{d}y }{\mathrm{d} x}$ at the point (1, 4) for the following equation:

.

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Answer

Using implicit differentiation, taking the derivative of the given equation yields

$y + \frac{\mathrm{d} y}{\mathrm{d} x}x + 6x -4y\frac{\mathrm{d} y}{\mathrm{d} x} = 0$

Getting all the $\frac{\mathrm{d} y}{\mathrm{d} x}$ 's to their own side, we have

$x\frac{\mathrm{d} y}{\mathrm{d} x} - 4y\frac{\mathrm{d} y}{\mathrm{d} x} = -6x - y$

Factoring,

$\frac{\mathrm{d} y}{\mathrm{d} x}(x - 4y) = -6x - y$

And dividing

$\frac{\mathrm{d}y }{\mathrm{d} x} = \frac{-6x - y}{x - 4y}$

Plugging in our point, , we have

$\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{-6 - 4}{1 - 16} = \frac{-10}{-15} =\frac{2}{3}$

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Question

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

$y^3+\sec(y)=3x^2$

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Answer

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

Taking the derivative with respect to x of both sides of the equation, we get

$3y^2 \frac{\mathrm{d} y}{\mathrm{d} x}+\sec(y)\tan(y)\frac{\mathrm{d} y}{\mathrm{d} x}=6x$

The derivatives were found using the following rules:

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

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 , so to speak, for the function it appears in.

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

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{6x}{3y^2+\sec(y)\tan(y)}$

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Question

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

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Answer

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

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

which was found using the following rules:

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

Note that 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 , so to speak, for the function it appears in.

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

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

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Question

Given that , compute the derivative of the following function

$y=6\sin(xy)+e^y$

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Answer

To find the derivative of the function, we use implicit differentiation, which is an application of the chain rule. We use this because , 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

$y'=6\cos(xy)(xy'+y)+y'e^y$

Then we expand in order to isolate the terms with

$y'=6xy'\cos(xy)+6y\cos(xy)+y'e^y$

$y'-6xy'\cos(xy)-y'e^y=6y\cos(xy)$

Then we factor out a

$y'(1-6x\cos(xy)-e^y)=6y\cos(xy)$

$y'=\frac{6y\cos(xy)}{1-6x\cos(xy)-e^y}$

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Question

Given that , compute the derivative of the following function:

$y=xy^2+4\sin(y)$

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Answer

To find the derivative of the function, we use implicit differentiation, which is an application of the chain rule. We use this because , 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

$y'=y^2+x(2y)(y')+4\cos(y)(y')$

Then isolate the terms with

$y'-2xyy'-4y'\cos(y)=y^2$

Then we factor out a

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

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

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Question

Given that , compute the derivative of the following function:

$y=\tan(xy^2)$

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Answer

To find the derivative of the function, we use implicit differentiation, which is an application of the chain rule. We use this because , 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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