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

Mar 6, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = - y {e}^{x} \frac{6 + 2 {e}^{y}}{1 + 6 y {e}^{y} + 2 y {e}^{2 y} - 2 y {e}^{x + y}}$

#### Explanation:

Differentiate both sides with respect to one variable, for ease's sake I'll differentiate with respect to $x$.

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

Assume the other variable is dependant on the one you're differentiating and use the chain rule to be able to differentiate them

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

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

For the middle one, say ${e}^{x} - {e}^{y} = u$ so $\frac{\mathrm{du}}{\mathrm{dx}} = {e}^{x} - {e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

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

Use algebra to isolate the derivate of the other variable

0 = 1/(y(e^x-e^y))dy/dx +e^x ln(y)/(e^x-e^y)^2 -e^yln(y)/(e^x-e^y)^2 dy/dx - e^x -e^ydy/dx

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

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

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

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

From there it's just a matter of simplifying, if you took $\frac{\mathrm{dx}}{\mathrm{dy}}$ and need $\frac{\mathrm{dy}}{\mathrm{dx}}$ just flip the denominator and numerator.

Dividing by $\left({e}^{x} - {e}^{y}\right)$ in top and below

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

From the original statemement we know that

$6 = \ln \frac{y}{{e}^{x} - {e}^{y}} - {e}^{x} - {e}^{y}$
$6 + {e}^{y} = \ln \frac{y}{{e}^{x} - {e}^{y}} - {e}^{x}$
$- 6 - {e}^{y} = {e}^{x} - \ln \frac{y}{{e}^{x} - {e}^{y}}$ and $6 + {e}^{y} - {e}^{x} = \ln \frac{y}{{e}^{x} - {e}^{y}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = y {e}^{x} \frac{- 6 - 2 {e}^{y}}{1 + y {e}^{y} \left(6 + {e}^{y} - {e}^{x}\right) - y {e}^{y} \left({e}^{x} - {e}^{y}\right)}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = y {e}^{x} \frac{- 6 - 2 {e}^{y}}{1 + 6 y {e}^{y} + y {e}^{2 y} - y {e}^{x + y} - y {e}^{x + y} + y {e}^{2 y}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = y {e}^{x} \frac{- 6 - 2 {e}^{y}}{1 + 6 y {e}^{y} + 2 y {e}^{2 y} - 2 y {e}^{x + y}}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = - y {e}^{x} \frac{6 + 2 {e}^{y}}{1 + 6 y {e}^{y} + 2 y {e}^{2 y} - 2 y {e}^{x + y}}$