# How do you differentiate e^(x/y)=x-y?

Jul 27, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - \left(x - y\right) \ln \left(x - y\right)}{y} , \mathmr{and} , \frac{x}{y} \left\{1 - \frac{1}{y} \cdot {e}^{\frac{x}{y}}\right\}$.

#### Explanation:

Prerequisites : The Usual Rules of Differentiation.

Given that, ${e}^{\frac{x}{y}} = x - y$.

$\therefore \ln {e}^{\frac{x}{y}} = \ln \left(x - y\right)$.

$\therefore \frac{x}{y} = \ln \left(x - y\right) , \mathmr{and} , x = y \ln \left(x - y\right)$.

$\therefore \frac{d}{\mathrm{dx}} \left\{x\right\} = \frac{d}{\mathrm{dx}} \left\{x \ln \left(x - y\right)\right\}$.

$\therefore 1 = x \frac{d}{\mathrm{dx}} \left\{\ln \left(x - y\right)\right\} + \ln \left(x - y\right) \cdot \frac{d}{\mathrm{dx}} \left\{y\right\} ,$

$i . e . , 1 = x \cdot \frac{1}{x - y} \cdot \frac{d}{\mathrm{dx}} \left\{x - y\right\} + \ln \left(x - y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} ,$

$\mathmr{and} , 1 = \frac{x}{x - y} \left\{\frac{d}{\mathrm{dx}} \left(x\right) - \frac{d}{\mathrm{dx}} \left(y\right)\right\} + \ln \left(x - y\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$.

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

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

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

$\therefore \frac{\left(x - y\right) - x}{\cancel{x - y}} = \frac{\left(x - y\right) \ln \left(x - y\right) - x}{\cancel{x - y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$.

$\therefore - y = \left\{\left(x - y\right) \ln \left(x - y\right) - x\right\} \frac{\mathrm{dy}}{\mathrm{dx}}$.

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x - \left(x - y\right) \ln \left(x - y\right)}{y}$.

Since, $\left(x - y\right) = {e}^{\frac{x}{y}} \mathmr{and} \ln \left(x - y\right) = \frac{x}{y}$, we may write,

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{x}{y} - \frac{1}{y} \cdot {e}^{\frac{x}{y}} \cdot \frac{x}{y} = \frac{x}{y} \left\{1 - \frac{1}{y} \cdot {e}^{\frac{x}{y}}\right\}$.