# How do you find the derivative of 2xy=y^2?

Mar 23, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{y - x}$ which can also be written as $\frac{y}{x}$.

#### Explanation:

Differentiating implicitly gets us

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

So that, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{y - x}$

Further discussion

$2 x y = {y}^{2}$ is equivalent to ${y}^{2} - 2 x y = 0$ which in turn is equivalent to $y \left(y - 2 x\right) = 0$.

The graph of this equation consists of the line $y = 0$ (the $x$ axis) and the line $y = 2 x$

graph{y(y-2x)=0 [-17.19, 14.86, -9.43, 6.59]}

Description of Cases
At $\left(0 , 0\right)$ there is no derivative.

For points $\left(x , 0\right)$ with $x \ne 0$, we have $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$.

Other points on the graph lie on $y = 2 x$ alone and have derivative $\frac{\mathrm{dy}}{\mathrm{dx}} = 2$

Rewriting the derivative

Our result above says $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{y - x}$.

For $y \ne 0$, we can write $\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} / \left({y}^{2} - x y\right)$

Since ${y}^{2} = 2 x y$ (see the original equation), we can write

$\frac{\mathrm{dy}}{\mathrm{dx}} = {y}^{2} / \left(2 x y - x y\right) = {y}^{2} / \left(x y\right) = \frac{y}{x}$

Checking the three cases, we see that

At $\left(0 , 0\right)$ we have $\frac{y}{x}$ does not exist..

For points $\left(x , 0\right)$ with $x \ne 0$, we have $\frac{y}{x} = 0$.

Other points on the graph have $y = 2 x$ alone and have derivative $\frac{y}{x} = \frac{2 x}{x} = 2$.

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x}$ is also true.