# How do you differentiate xe^y=y-1?

Oct 9, 2016

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

#### Explanation:

Generally differentiation problems involve functions i.e. $y = f \left(x\right)$ - written explicitly as functions of $x$. However, some functions of $y$ are written implicitly as functions of $x$, like the one in question above.

So what we do is to treat $y$ as $y = y \left(x\right)$ and use chain rule. This means differentiating $y$ w.r.t. $y$, but as we have to derive w.r.t. $x$, as per chain rule, we multiply it by $\frac{\mathrm{dy}}{\mathrm{dx}}$.

Using this, differential of $x {e}^{y} = y - 1$ can be derived as follows:

$x \times \frac{d}{\mathrm{dy}} \cdot {e}^{y} \times \frac{\mathrm{dy}}{\mathrm{dx}} + 1 \times {e}^{y} = \frac{d}{\mathrm{dy}} y \times \frac{\mathrm{dy}}{\mathrm{dx}}$

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

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

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

Oct 9, 2016

${e}^{y} / \left(2 - y\right)$.

#### Explanation:

#x=e^(-y)(y-1) is from the given solvable-for-x implicit form.

In such problems, use

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = y ' = {e}^{y} / \left(2 - y\right)$..