# How do you differentiate -3=xye^(x-y)?

Dec 2, 2015

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

#### Explanation:

To find the implicit derivative, we'll use the chain rule and product rule. It's important to remember that when differentiating a $y$ term, a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term will be spit out thanks to the chain rule.

$\frac{d}{\mathrm{dx}} \left[- 3 = x y {e}^{x - y}\right]$

$0 = y {e}^{x - y} \frac{d}{\mathrm{dx}} \left[x\right] + x {e}^{x - y} \frac{d}{\mathrm{dx}} \left[y\right] + x y \frac{d}{\mathrm{dx}} \left[{e}^{x - y}\right]$

Find each derivative.

$\frac{d}{\mathrm{dx}} \left[x\right] = 1$

$\frac{d}{\mathrm{dx}} \left[y\right] = \frac{\mathrm{dy}}{\mathrm{dx}}$

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

Plug them back in and solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

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

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

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

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

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