# How do you use Implicit Differentiation of ye^x=xe^y?

Mar 8, 2018

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

#### Explanation:

First we take $\frac{d}{\mathrm{dx}}$ of each term.

$\frac{d}{\mathrm{dx}} \left[y {e}^{x}\right] = \frac{d}{\mathrm{dx}} \left[x {e}^{y}\right]$

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

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

Using the chain rule, we know that:
$\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$y {e}^{x} + \frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} \frac{d}{\mathrm{dy}} \left[y\right] = \frac{\mathrm{dy}}{\mathrm{dx}} x \frac{d}{\mathrm{dy}} \left[{e}^{y}\right] + {e}^{y}$

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

Now gather like terms together.
$\frac{\mathrm{dy}}{\mathrm{dx}} {e}^{x} - \frac{\mathrm{dy}}{\mathrm{dx}} x {e}^{y} = {e}^{y} - y {e}^{x}$

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

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