# What is the slope of f(x)=-X/e^(x-x^3)  at x=-1?

Slope $= + 1$

#### Explanation:

Given $f \left(x\right) = - \frac{x}{{e}^{x - {x}^{3}}}$ at $x = - 1$

Determine the first derivative $f ' \left(x\right)$ first, then compute for $f ' \left(- 1\right)$ to obtain the slope

We make use of the derivative of quotient for this type of function

use the formula $\frac{d}{\mathrm{dx}} \left(\frac{u}{v}\right) = \frac{v \cdot \frac{d}{\mathrm{dx}} \left(u\right) - u \cdot \frac{d}{\mathrm{dx}} \left(v\right)}{v} ^ 2$

Let $u = x$ and $v = {e}^{x - {x}^{3}}$

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) =$
$\frac{d}{\mathrm{dx}} \left(- \frac{x}{{e}^{x - {x}^{3}}}\right) = \left(- 1\right) \cdot \frac{{e}^{x - {x}^{3}} \cdot \frac{d}{\mathrm{dx}} \left(x\right) - x \cdot \frac{d}{\mathrm{dx}} \left({e}^{x - {x}^{3}}\right)}{{e}^{x - {x}^{3}}} ^ 2$

$\frac{d}{\mathrm{dx}} \left(- \frac{x}{{e}^{x - {x}^{3}}}\right) = \left(- 1\right) \cdot \frac{\left({e}^{x - {x}^{3}} \cdot 1 - x {e}^{x - {x}^{3}} \left(1 - 3 {x}^{2}\right)\right)}{{e}^{x - {x}^{3}}} ^ 2$

There is no need for further simplification just use $x = - 1$ right away

$f ' \left(- 1\right) = \left(- 1\right) \cdot \frac{\left({e}^{- 1 - {\left(- 1\right)}^{3}} \cdot 1 - \left(- 1\right) {e}^{- 1 - {\left(- 1\right)}^{3}} \left(1 - 3 {\left(- 1\right)}^{2}\right)\right)}{{e}^{- 1 - {\left(- 1\right)}^{3}}} ^ 2$

$f ' \left(- 1\right) = \left(- 1\right) \cdot \frac{\left({e}^{- 1 + 1} - \left(- 1\right) {e}^{- 1 + 1} \left(1 - 3\right)\right)}{{e}^{- 1 + 1}} ^ 2$

$f ' \left(- 1\right) = \left(- 1\right) \cdot \frac{\left({e}^{0} - \left(- 1\right) {e}^{0} \left(- 2\right)\right)}{{e}^{0}} ^ 2$

$f ' \left(- 1\right) = \left(- 1\right) \cdot \frac{\left(1 - 2\right)}{1} ^ 2$

$f ' \left(- 1\right) = + 1$

God bless....I hope the explanation is useful.