# How do you implicitly differentiate 7=1-e^y/(xy)?

Apr 18, 2016

I found: $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y}{x \left(y - 1\right)}$

#### Explanation:

We need to remember that $y$ will be itself a function of $x$ so we need to derive it accordingly.
For example, if you have ${y}^{2}$ the derivative will be:
$2 y \cdot \frac{\mathrm{dy}}{\mathrm{dx}}$
to consider its dependence from $x$.
In our case we have:
$0 = 0 - \frac{x y {e}^{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{y} \left(y + x \frac{\mathrm{dy}}{\mathrm{dx}}\right)}{x y} ^ 2$
rearranging:
$\frac{\mathrm{dy}}{\mathrm{dx}} \left[x y {e}^{y} - x {e}^{y}\right] = y {e}^{y}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{y {e}^{y}}{x {e}^{y} \left(y - 1\right)} = \frac{y}{x \left(y - 1\right)}$