# How do you implicitly differentiate 7=-x+(xe^y)/(x-y)?

Nov 19, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} + 7 y}{x \left(x + 7\right) \left(1 - x + y\right)}$

#### Explanation:

We have

$7 = - x + \frac{x {e}^{y}}{x - y}$

Arranging that we have

$7 + x = \frac{x {e}^{y}}{x - y}$

$x - y = \frac{x {e}^{y}}{x + 7}$

$y = x - \frac{x {e}^{y}}{x + 7}$

Derivating we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} x - x {e}^{y} \frac{d}{\mathrm{dx}} \frac{1}{x + 7} - \frac{1}{x + 7} \frac{d}{\mathrm{dx}} x {e}^{y}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{x {e}^{y}}{x + 7} ^ 2 - \frac{1}{x + 7} \left({e}^{y} \frac{d}{\mathrm{dx}} x + x \frac{d}{\mathrm{dx}} {e}^{y}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{x {e}^{y}}{x + 7} ^ 2 - \frac{1}{x + 7} \left({e}^{y} + x \frac{d}{\mathrm{dy}} {e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{x {e}^{y}}{x + 7} ^ 2 - \frac{1}{x + 7} \left({e}^{y} + x {e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{x {e}^{y}}{x + 7} ^ 2 - {e}^{y} / \left(x + 7\right) + \frac{x {e}^{y}}{x + 7} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} - \frac{x {e}^{y}}{x + 7} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = 1 + \frac{x {e}^{y}}{x + 7} ^ 2 - {e}^{y} / \left(x + 7\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(1 - \frac{x {e}^{y}}{x + 7}\right) = 1 + \frac{x {e}^{y}}{x + 7} ^ 2 - {e}^{y} / \left(x + 7\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(1 - \frac{x {e}^{y}}{x + 7}\right)}^{- 1} \left(1 + \frac{x {e}^{y}}{x + 7} ^ 2 - {e}^{y} / \left(x + 7\right)\right)$

However, $x - y = \frac{x {e}^{y}}{x + 7}$, so we can rewrite this to

$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(1 - x + y\right)}^{- 1} \left(1 + \frac{x - y}{x + 7} - \frac{x - y}{x}\right)$

After summing those fractions we reach

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} + 7 y}{x \left(x + 7\right) \left(1 - x + y\right)}$