# How do you find the implicit differentiation of xe^y+ye^x=1?

Nov 30, 2016

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

#### Explanation:

differentiate $\textcolor{b l u e}{\text{implicitly with respect to x}}$

Both terms on the left have to be differentiated using the $\textcolor{b l u e}{\text{product rule}}$

$\Rightarrow \left(x {e}^{y} . \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} .1\right) + \left(y {e}^{x} + {e}^{x} . \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

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

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