# How do you implicitly differentiate y = e^x y -x e^y?

Jan 9, 2016

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

#### Explanation:

We consider $y$ as a function of $x$ and derive it as well to get:
$1 \frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x} y + {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} - x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[{e}^{x} - x {e}^{y} - 1\right] = {e}^{y} - {e}^{x} y$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{x} y - {e}^{y}}{1 + x {e}^{y} - {e}^{x}}$

Jan 9, 2016

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

#### Explanation:

differentiate using chain rule and product rule.

Differentiating with respect to x gives :

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

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

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

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

collecting$\frac{\mathrm{dy}}{\mathrm{dx}}$ terms to left hand side gives :

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

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

 rArr dy/dx = (ye^x - e^y )/(1 - e^x + xe^y