# What is the slope of the tangent line of  (y+e^x)/(y-e^y) =C , where C is an arbitrary constant, at (-2,1)?

##### 1 Answer
Feb 3, 2017

The slope at $\left(2 , 1\right)$ is 1

#### Explanation:

$\frac{y + {e}^{x}}{y - {e}^{y}} = C$ is a family of equations, and we are specifically interested in those family members that pass through $\left(2 , 1\right)$.

One approach is to proceed with the calculus, as this is posted as a claculus problem. We can do some simplifying algebra first:

$y + {e}^{x} = C \left(y - {e}^{y}\right) \implies y \left(1 - C\right) = - {e}^{x} - C {e}^{y}$

If we now differentiate wrt x in order to obtain the slope:

$y ' \left(1 - C\right) = - {e}^{x} - y ' C {e}^{y}$

$\implies y ' \left(1 + C \left({e}^{y} - 1\right)\right) = - {e}^{x}$

$\implies y ' = - {e}^{x} / \left(1 + C \left({e}^{y} - 1\right)\right)$, the slope

At $\left(- 2 , 1\right)$ that means that:

$y ' = - \frac{1}{{e}^{2} \left(1 + C \left(e - 1\right)\right)}$

But we also know that $\frac{1 + {e}^{- 2}}{1 - e} = C$ and so we can sub that into our slope equation to obtain:

$y ' = - \frac{1}{{e}^{2} \left(1 + \frac{1 + {e}^{- 2}}{1 - e} \left(e - 1\right)\right)}$

$= - \frac{1}{{e}^{2} \left(1 - \left(1 + {e}^{- 2}\right)\right)}$

$= - \frac{1}{- 1} \textcolor{b l u e}{= 1}$

This is how it occurred me to do it off the bat. I reckon there must be better/cleverer ways but I leave that to you!