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

##### 1 Answer
Aug 6, 2016

The slope is $\frac{1}{e} ^ 3$.

#### Explanation:

This will be easier rewritten as:

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

Now, we can use implicit differentiation (remember that each occurrence of $y$ will put the chain rule in effect):

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

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

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

So, we want to solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$ when $\left(x , y\right) = \left(- 2 , 1\right)$.

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

Cross-multiplying:

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{e} ^ 3$