# How do you use implicit differentiation to find dy/dx given xe^y-y=5?

##### 1 Answer
Mar 5, 2017

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

#### Explanation:

When we differentiate $y$ wrt $x$ we get $\frac{\mathrm{dy}}{\mathrm{dx}}$.

However, we cannot differentiate a non implicit function of $y$ wrt $x$. But if we apply the chain rule we can differentiate a function of $y$ wrt $y$ but we must also multiply the result by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

When this is done in situ it is known as implicit differentiation.

We have:

$x {e}^{y} - y = 5$

Differentiate wrt $x$ (applying product rule):

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

Advanced Calculus

There is another (often faster) approach using partial derivatives. Suppose we cannot find $y$ explicitly as a function of $x$, only implicitly through the equation $F \left(x , y\right) = 0$ which defines $y$ as a function of $x , y = y \left(x\right)$. Therefore we can write $F \left(x , y\right) = 0$ as $F \left(x , y \left(x\right)\right) = 0$. Differentiating both sides of this, using the partial chain rule gives us

 (partial F)/(partial x) (1) + (partial F)/(partial y) dy/dx = 0 => dy/dx = −((partial F)/(partial x)) / ((partial F)/(partial y))

So Let $F \left(x , y\right) = x {e}^{y} - y - 5$; Then;

$\frac{\partial F}{\partial x} = {e}^{y}$

$\frac{\partial F}{\partial y} = x {e}^{y} - 1$

And so:

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{{e}^{y}}{x {e}^{y} - 1} = \frac{{e}^{y}}{1 - x {e}^{y}}$, as before.