# How do you find the derivative of xe^y+y=xy?

Apr 12, 2018

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

#### Explanation:

$x {e}^{y} + y = x y$.

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

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

Here, by the Chain Rule, $\frac{d}{\mathrm{dx}} \left({e}^{y}\right) = \frac{d}{\mathrm{dy}} \left({e}^{y}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$.

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

$\therefore x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} - x \frac{\mathrm{dy}}{\mathrm{dx}} = y - {e}^{y} , \mathmr{and} ,$

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

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

Feel the Joy of Maths.!