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

Nov 24, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x}{2 y - 1}$

#### Explanation:

Note that I'm assuming we are differentiating implicitly with respect to $x$.

$\frac{d}{\mathrm{dx}} \left[1 = {x}^{2} - y + {y}^{2}\right]$

Remember that any term that's not an $x$ (in this case, $y$) will put chain rule into effect and spit out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term.

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

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

$- 2 x = \left(2 y - 1\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{2 x}{2 y - 1}$