# How do you implicitly differentiate -1=y^2+(xy-e^y)/(x)?

##### 1 Answer
Dec 24, 2015

The key to answering implicit differentiation problems is to derive portions of the equation involving a $y$ the same as you would the portions with $x$, remembering to multiply by a $\frac{\mathrm{dy}}{\mathrm{dx}}$, which you will later isolate to solve.

#### Explanation:

Rewriting the problem,

$- 1 = {y}^{2} + \frac{x y}{x} - \frac{{e}^{y}}{x}$
$- 1 = {y}^{2} + y - \frac{{e}^{y}}{x}$

We now derive both sides of the equation, remembering to include a $\frac{\mathrm{dy}}{\mathrm{dx}}$ when deriving any functions of $y$:

$\frac{d}{\mathrm{dx}} \left(- 1\right) = \frac{d}{\mathrm{dx}} \left({y}^{2} + y - {e}^{y} {x}^{-} 1\right)$
$0 = 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} - \left({e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} {x}^{-} 1 + \left(- 1\right) {e}^{y} {x}^{-} 2\right)$
$0 = 2 y \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} {x}^{-} 1 \frac{\mathrm{dy}}{\mathrm{dx}} + {e}^{y} {x}^{-} 2$

We now move all terms with $\frac{\mathrm{dy}}{\mathrm{dx}}$ to one side of the equations in order to isolate the term:

$2 y \frac{\mathrm{dy}}{\mathrm{dx}} + \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} {x}^{-} 1 \frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{y} {x}^{-} 2$

We can now isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 y + 1 - {e}^{y} {x}^{-} 1\right) = - {e}^{y} {x}^{-} 2$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {e}^{y} {x}^{-} 2}{2 y + 1 - {e}^{y} {x}^{-} 1}$

The $x$ terms with negative powers can now be rewritten to simplify:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{- {e}^{y}}{x} ^ 2}{2 y + 1 - {e}^{y} / x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- {e}^{y}}{2 y {x}^{2} + {x}^{2} - {e}^{y} x}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{y}}{{e}^{y} x - {x}^{2} - 2 y {x}^{2}}$

Giving us the simplified answer:

dy/dx = (e^y) / (x(e^y - x[1 + 2y])