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

Jun 28, 2016

The reqd. slope $= {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{x = - 1} = f ' \left(- 1\right) = - 9 {e}^{2.}$

#### Explanation:

Slope of a curve $: y = f \left(x\right)$ at a particular pt. is, by defn., the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ at that pt.

Here, $y = f \left(x\right) = - x \cdot {e}^{x - 3 {x}^{3}} .$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = f ' \left(x\right) = - \left\{x \cdot \frac{d}{\mathrm{dx}} \left({e}^{x - 3 {x}^{3}}\right) + {e}^{x - 3 {x}^{3}} \cdot \frac{d}{\mathrm{dx}} \left(x\right)\right\} = - \left\{x \cdot {e}^{x - 3 {x}^{3}} \cdot \frac{d}{\mathrm{dx}} \left(\left(x - 3 {x}^{3}\right)\right) + {e}^{x - 3 {x}^{3}} \cdot 1\right\} = - \left\{x \cdot {e}^{x - 3 {x}^{3}} \cdot \left(1 - 9 {x}^{2}\right) + {e}^{x - 3 {x}^{3}}\right\} = - {e}^{x - 3 {x}^{3}} \left\{x - 9 {x}^{3} + 1\right\} .$

Therefore, the reqd. slope $= {\left[\frac{\mathrm{dy}}{\mathrm{dx}}\right]}_{x = - 1} = f ' \left(- 1\right) = - {e}^{- 1 + 3} \left(- 1 + 9 + 1\right) = - 9 {e}^{2}$