# What is the slope of f(x)=-e^x/(x-2 at x=-2?

Jul 17, 2017

$f ' \left(- 2\right) = \frac{5}{16 {e}^{2}}$

#### Explanation:

We have: $f \left(x\right) = - \frac{{e}^{x}}{x - 2}$

First, let's find the function for the slope, $f ' \left(x\right)$, by differentiating $f \left(x\right)$:

$R i g h t a r r o w f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(- \frac{{e}^{x}}{x - 2}\right)$

$R i g h t a r r o w f ' \left(x\right) = - \frac{d}{\mathrm{dx}} \left(\frac{{e}^{x}}{x - 2}\right)$

Using the quotient law of differentiation:

$R i g h t a r r o w f ' \left(x\right) = - \frac{\left(x - 2\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right) - \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \left(x - 2\right)}{{\left(x - 2\right)}^{2}}$

$R i g h t a r r o w f ' \left(x\right) = - \frac{\left(x - 2\right) \cdot {e}^{x} - \left({e}^{x}\right) \cdot 1}{{\left(x - 2\right)}^{2}}$

$R i g h t a r r o w f ' \left(x\right) = - \frac{x {e}^{x} - 2 {e}^{x} - {e}^{x}}{{\left(x - 2\right)}^{2}}$

$R i g h t a r r o w f ' \left(x\right) = - \frac{x {e}^{x} - 3 {e}^{x}}{{\left(x - 2\right)}^{2}}$

$\therefore f ' \left(x\right) = - \frac{{e}^{x} \left(x - 3\right)}{{\left(x - 2\right)}^{2}}$

Now, we need to evaluate the slope at $x = - 2$.

So let's substitute $- 2$ in place of $x$:

$R i g h t a r r o w f ' \left(- 2\right) = - \frac{{e}^{\left(- 2\right)} \left(\left(- 2\right) - 3\right)}{{\left(\left(- 2\right) - 2\right)}^{2}}$

$R i g h t a r r o w f ' \left(- 2\right) = - \frac{{e}^{- 2} \left(- 5\right)}{{\left(- 4\right)}^{2}}$

$R i g h t a r r o w f ' \left(- 2\right) = - \frac{- \frac{5}{{e}^{2}}}{16}$

$\therefore f ' \left(- 2\right) = \frac{5}{16 {e}^{2}}$

Therefore, at $x = - 2$, the slope of $f \left(x\right)$ is $\frac{5}{16 {e}^{2}}$.