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# Implicit Differentiation Calculator

## Get detailed solutions to your math problems with our Implicit Differentiation step-by-step calculator. Practice your math skills and learn step by step with our math solver. Check out all of our online calculators here.

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###  Difficult Problems

1

Here, we show you a step-by-step solved example of implicit differentiation. This solution was automatically generated by our smart calculator:

$\frac{d}{dx}\left(x^2+y^2=16\right)$
2

Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(16\right)$
3

The derivative of the constant function ($16$) is equal to zero

$\frac{d}{dx}\left(x^2+y^2\right)=0$
4

The derivative of a sum of two or more functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)=0$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{d}{dx}\left(x^2\right)+2y^{2-1}\frac{d}{dx}\left(y\right)=0$

Add the values $2$ and $-1$

$\frac{d}{dx}\left(x^2\right)+2y^{1}\frac{d}{dx}\left(y\right)=0$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$2y^{2-1}\frac{d}{dx}\left(y\right)$

Subtract the values $2$ and $-1$

$2y^{1}\frac{d}{dx}\left(y\right)$
5

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$\frac{d}{dx}\left(x^2\right)+2y^{1}\frac{d}{dx}\left(y\right)=0$
6

Any expression to the power of $1$ is equal to that same expression

$\frac{d}{dx}\left(x^2\right)+2y\frac{d}{dx}\left(y\right)=0$
7

The derivative of the linear function is equal to $1$

$\frac{d}{dx}\left(x^2\right)+2y\cdot y^{\prime}=0$

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$2x^{\left(2-1\right)}$

Subtract the values $2$ and $-1$

$2x$
8

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$2x+2y\cdot y^{\prime}=0$
9

We need to isolate the dependent variable $y$, we can do that by simultaneously subtracting $2x$ from both sides of the equation

$2y\cdot y^{\prime}=-2x$
10

Divide both sides of the equation by $2$

$y^{\prime}y=\frac{-2x}{2}$
11

Take $\frac{-2}{2}$ out of the fraction

$y^{\prime}y=-x$
12

Divide both sides of the equation by $y$

$y^{\prime}=\frac{-x}{y}$

##  Final answer to the problem

$y^{\prime}=\frac{-x}{y}$

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