# What is the slope of the tangent line of 1/(e^y-e^x) = C , where C is an arbitrary constant, at (-1,1)?

Jan 6, 2016

The slope is $\frac{1}{e} ^ 2 = {e}^{- 2}$ and $C$ is not arbitrary, it is $\frac{e}{{e}^{2} - 1}$.

#### Explanation:

Given $\frac{1}{{e}^{y} - {e}^{x}} = C$ for constant $C$.

We can rewrite this as ${e}^{y} - {e}^{x} = \frac{1}{C}$.

Now differentiate implicitly.

${e}^{y} \frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x} = 0$. So,

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

Given that the point $\left(- 1 , 1\right)$ lies on the graph of this equation, we have

dy/dx ]_"(-1,1)" = e^(-1)/e^1 = 1/e^2

And $\left(- 1 , 1\right)$ being on the graph also determines $C$ for this graph.
Since $\frac{1}{{e}^{y} - {e}^{x}} = C$, we find that $C = \frac{1}{e - \frac{1}{e}} = \frac{e}{{e}^{2} - 1}$.