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

Feb 24, 2017

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

#### Explanation:

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

$\rightarrow \textcolor{w h i t e}{\text{XX}} {y}^{2} - x y = 2$

$\rightarrow \textcolor{w h i t e}{\text{XX}} \frac{d \left({y}^{2} - x y\right)}{\mathrm{dx}} = \frac{d \left(2\right)}{\mathrm{dx}}$

$\rightarrow \textcolor{w h i t e}{\text{XX}} \frac{d \left({y}^{2}\right)}{\mathrm{dx}} - \frac{d \left(x y\right)}{\mathrm{dx}} = 0$

$\rightarrow \textcolor{w h i t e}{\text{XX}} 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$
$\textcolor{w h i t e}{\text{XXXXXXXXXXXX}}$(using the Chain and Product rules)

$\rightarrow \textcolor{w h i t e}{\text{XX}} 2 y \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} = y$

$\rightarrow \textcolor{w h i t e}{\text{XX}} \frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y - x\right) = y$

$\rightarrow \textcolor{w h i t e}{\text{XX}} \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{2 y - x}$