Find the derivative of yx^2+xy^2=0+-1*211111

\frac{d}{dx}\left(yx^2+xy^2=0-1\cdot 211111\right)

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$y^2+2y\cdot x=0$

Step by step solution

Problem

$\frac{d}{dx}\left(yx^2+xy^2=0-1\cdot 211111\right)$
1

Subtract the values $0$ and $-211111$

$\frac{d}{dx}\left(xy^2+yx^2=-211111\right)$
2

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

$\frac{d}{dx}\left(xy^2+yx^2\right)=\frac{d}{dx}\left(-211111\right)$
3

The derivative of the constant function is equal to zero

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

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

$\frac{d}{dx}\left(xy^2\right)+\frac{d}{dx}\left(yx^2\right)=0$
5

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

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

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

$1y^2+y\frac{d}{dx}\left(x^2\right)=0$
7

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

$1y^2+2y\cdot x=0$
8

Any expression multiplied by $1$ is equal to itself

$y^2+2y\cdot x=0$

$y^2+2y\cdot x=0$

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Main topic:

Differential calculus

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