# How do you implicitly differentiate y= y^2 + e^(x y) ?

Dec 1, 2016

We use the power rule and the rule if y= e^(f(x), then its derivative $y ' = f ' \left(x\right) {e}^{f \left(x\right)}$.

Hence:

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

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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

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

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

Hopefully this helps!