# How do you use implicit differentiation to find (dy)/(dx) given 2xy-y^2=3?

##### 1 Answer
Oct 9, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{y - 1}$

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{2 x y} - \textcolor{b l u e}{{y}^{2}} = \textcolor{g r e e n}{3}$

Note 1:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{\frac{d \left(2 x y\right)}{\mathrm{dx}}} = 2 x \cdot \frac{\mathrm{dy}}{\mathrm{dx}} + y \cdot \frac{d \left(2 x\right)}{\mathrm{dx}} = \textcolor{red}{2 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y}$

Note 2:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{\frac{d \left({y}^{2}\right)}{\mathrm{dx}}} = \textcolor{b l u e}{2 y \frac{\mathrm{dy}}{\mathrm{dx}}}$

Note 3:
$\textcolor{w h i t e}{\text{XXX}} \textcolor{g r e e n}{\frac{d \left(3\right)}{\mathrm{dx}}} = \textcolor{g r e e n}{0}$

Therefore $2 x y - {y}^{2} = 3$

$\Rightarrow \textcolor{red}{2 \frac{\mathrm{dy}}{\mathrm{dx}} + 2 y} - \textcolor{b l u e}{2 y \frac{\mathrm{dy}}{\mathrm{dx}}} = \textcolor{g r e e n}{0}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} - y \frac{\mathrm{dy}}{\mathrm{dx}} = - y$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - y\right) = - y$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{y - 1}$