# How do you implicitly differentiate 9=e^y/e^x-xy?

Jan 7, 2016

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

#### Explanation:

Before we start, I'd like to simplify the given equation.

$9 = {e}^{y - x} - x y$, this may make things easier.

First step in implicit differentiation is to apply the derivative operator to both sides.

$\frac{d}{\mathrm{dx}} \left[9\right] = \frac{d}{\mathrm{dx}} \left[{e}^{y - x} - x y\right]$, the left hand side is trivial, namely, it's just $0$. By applying the chain rule to ${e}^{y - x}$ and the product rule to $x y$, we obtain the following:

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

$0 = {e}^{y - x} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y - x} - y + x \frac{\mathrm{dy}}{\mathrm{dx}}$

Doing a bit of algebra, and factorising that $\frac{\mathrm{dy}}{\mathrm{dx}}$, we get,

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

Therefore, $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{e}^{y - x} + y}{{e}^{y - x} + x}$.

I suppose, to simplify further, we can multiply the numerator and the denominator by ${e}^{x}$ on the right hand side:

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