# How do you implicitly differentiate -1=yx+(x-y-e^y)/(y+x)?

Mar 30, 2017

It's simple.

#### Explanation:

$d \left(- 1\right) = 0$
$d \left(x y\right) = \mathrm{dx} y + x \mathrm{dy}$
$d {\left(y + x\right)}^{- 1} = - \frac{1}{y + x} ^ 2 \left(\mathrm{dx} + \mathrm{dy}\right)$
$d \left(x - y - {e}^{y}\right) = \mathrm{dx} - \mathrm{dy} - \mathrm{dy} {e}^{y}$
so
$0 = y \mathrm{dx} + x \mathrm{dy} - \left(x - y - {e}^{y}\right) \frac{1}{y + x} ^ 2 \left(\mathrm{dx} + \mathrm{dy}\right) + \frac{1}{y + x} \left(\mathrm{dx} - \mathrm{dy} - \mathrm{dy} {e}^{y}\right)$
You can continue and put all the dx and dy terms together and divide by dx. But that's going beyond what is asked, which is the differential, not the derivative.
$\mathrm{dy} \left(x + \frac{x - y - {e}^{y}}{y + x} ^ 2 - \frac{1 + {e}^{y}}{y + x}\right) = \mathrm{dx} \left(- y - \frac{x - y - {e}^{y}}{y + x} ^ 2 + \frac{1}{y + x}\right)$
This is only useful if you can separate variables.