# How do you differentiate 1=e^(xy)/(e^x+e^y)?

Jul 12, 2016

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

#### Explanation:

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

and
${e}^{x} + {e}^{y} - {e}^{x y} = 0$

then the implicit function theorem states that $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{f}_{x}}{{f}_{y}}$

${f}_{x} = {e}^{x} - y {e}^{x y}$

${f}_{y} = {e}^{y} - x {e}^{x y}$

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