# How do you implicitly differentiate 1=-y^2/x+xy?

Dec 12, 2015

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

#### Explanation:

First, we will rearrange the equation:

$1 = {y}^{2} / x = x y$

$1 + {y}^{2} = {x}^{2} y$

In the above part, we've just sended the x to the right side and multiplied it.

Now, differentiating w.r.to x

$0 + 2 y \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} + 2 x y$

On the right hand in the above equation, we've applied the product rule

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

Taking $\frac{\mathrm{dy}}{\mathrm{dx}}$ common:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left[2 y - {x}^{2}\right] = 2 x y$

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

The above equation will be our final answer.