# How do you implicitly differentiate 6=lny/(e^x-e^y)-x?

Sep 1, 2016

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

#### Explanation:

$\frac{d}{\mathrm{dx}} \left(6\right) = \frac{d}{\mathrm{dx}} \left(\frac{\ln y}{{e}^{x} - {e}^{y}} - x\right)$

$\frac{d}{\mathrm{dx}} \left(6\right) = \frac{d}{\mathrm{dx}} \left(\ln \frac{y}{{e}^{x} - {e}^{y}}\right) + \frac{d}{\mathrm{dx}} \left(- x\right)$

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

Isolate $\frac{\mathrm{dy}}{\mathrm{dx}}$:

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

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

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

Hopefully this helps!