# How do you differentiate 5y-x=(xy)/ln(x-y)?

Apr 14, 2016

$y ' = \frac{y \ln \left(x - y\right) - \frac{x y}{x - y} + {\left(\ln \left(x - y\right)\right)}^{2}}{5 {\left(\ln \left(x - y\right)\right)}^{2} - x \ln \left(x - y\right) - \frac{x y}{x - y}}$

#### Explanation:

$5 y ' - 1 = \frac{\ln \left(x - y\right) \left[x y ' + y\right] - x y \left(\frac{1}{x - y} \left(1 - y '\right)\right)}{\ln \left(x - y\right)} ^ 2$

(5y'-1)xx(ln(x-y))^2=ln(x-y)[xy'+y]-(xy)/(x-y)(1-y'))

$5 y ' {\left(\ln \left(x - y\right)\right)}^{2} - {\left(\ln \left(x - y\right)\right)}^{2} = x y ' \ln \left(x - y\right) + y \ln \left(x - y\right) - \frac{x y}{x - y} + \frac{x y y '}{x - y}$

$5 y ' {\left(\ln \left(x - y\right)\right)}^{2} - x y ' \ln \left(x - y\right) - \frac{x y y '}{x - y} = y \ln \left(x - y\right) - \frac{x y}{x - y} + {\left(\ln \left(x - y\right)\right)}^{2}$

$y ' \left[5 {\left(\ln \left(x - y\right)\right)}^{2} - x \ln \left(x - y\right) - \frac{x y}{x - y}\right] = y \ln \left(x - y\right) - \frac{x y}{x - y} + {\left(\ln \left(x - y\right)\right)}^{2}$

$y ' = \frac{y \ln \left(x - y\right) - \frac{x y}{x - y} + {\left(\ln \left(x - y\right)\right)}^{2}}{5 {\left(\ln \left(x - y\right)\right)}^{2} - x \ln \left(x - y\right) - \frac{x y}{x - y}}$