# How do you find the derivative of 7=xy-e^(xy)?

Sep 18, 2016

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

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(7\right) = \frac{d}{\mathrm{dx}} \left(x y - {e}^{x y}\right)$

$\frac{d}{\mathrm{dx}} \left(7\right) = \frac{d}{\mathrm{dx}} \left(x y\right) + \frac{d}{\mathrm{dx}} \left({e}^{x y}\right)$

By using the identity $\frac{d}{\mathrm{dx}} \left({e}^{b \left(x\right)}\right) = \left(b \left(x\right)\right) ' \times {e}^{b \left(x\right)}$

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

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

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

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

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

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

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

Hopefully this helps!