# How do you implicitly differentiate 11=(x-y)(e^y+e^x)?

##### 1 Answer
Apr 2, 2016

Use the power rule and some algebra to get $\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{e}^{y} + {e}^{x} + x {e}^{x} - y {e}^{x}}{x {e}^{y} - {e}^{y} - {e}^{x} - y {e}^{y}}$.

#### Explanation:

Begin by taking the derivative with respect to $x$ on both sides:
$\frac{d}{\mathrm{dx}} \left(11\right) = \frac{d}{\mathrm{dx}} \left(\left(x - y\right) \left({e}^{y} + {e}^{x}\right)\right)$

The left side is easy - the derivative of a constant is $0$:
$0 = \frac{d}{\mathrm{dx}} \left(\left(x - y\right) \left({e}^{y} + {e}^{x}\right)\right)$

The right side is a little harder. Because there's multiplication going on there, we'll have to apply the product rule, which says:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u ' v + u v '$

In our case,
$u = x - y \to u ' = 1 - \frac{\mathrm{dy}}{\mathrm{dx}}$
$v = {e}^{y} + {e}^{x} \to v ' = {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{x}$

Applying the power rule:
$\frac{d}{\mathrm{dx}} \left(\left(x - y\right) \left({e}^{y} + {e}^{x}\right)\right) = \left(x - y\right) ' \left({e}^{y} + {e}^{x}\right) + \left(x - y\right) \left({e}^{y} + {e}^{x}\right) '$
$\textcolor{w h i t e}{X X} = \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left({e}^{y} + {e}^{x}\right) + \left(x - y\right) \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{x}\right)$

Doing some algebra to isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$:
$0 = \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) \left({e}^{y} + {e}^{x}\right) + \left(x - y\right) \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{x}\right)$
$\textcolor{w h i t e}{X X} = {e}^{y} + {e}^{x} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} + x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} + x {e}^{x} - y {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - y {e}^{x}$
$\textcolor{w h i t e}{X X} = \left(x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} - y {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}\right) + \left({e}^{y} + {e}^{x} + x {e}^{x} - y {e}^{x}\right)$
$- \left({e}^{y} + {e}^{x} + x {e}^{x} - y {e}^{x}\right) = x {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x} \frac{\mathrm{dy}}{\mathrm{dx}} - y {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$
$- \left({e}^{y} + {e}^{x} + x {e}^{x} - y {e}^{x}\right) = \frac{\mathrm{dy}}{\mathrm{dx}} \left(x {e}^{y} - {e}^{y} - {e}^{x} - y {e}^{y}\right)$
$- \frac{{e}^{y} + {e}^{x} + x {e}^{x} - y {e}^{x}}{x {e}^{y} - {e}^{y} - {e}^{x} - y {e}^{y}} = \frac{\mathrm{dy}}{\mathrm{dx}}$

And there you have it.