# What is the slope of f(x)=-e^x/x^2 at x=1?

Sep 24, 2016

The Reqd. Slope$= e$.

#### Explanation:

We know that slope of a curve $C : y = f \left(x\right) = - {e}^{x} / {x}^{2} = - {x}^{-} 2 {e}^{x}$

at $x = 1 \text{ is nothing but } f ' \left(1\right) .$

$f \left(x\right) = - {x}^{-} 2 {e}^{x} \Rightarrow f ' \left(x\right) = - \left\{\left({x}^{-} 2\right) \left({e}^{x}\right) ' + \left({e}^{x}\right) \left({x}^{-} 2\right) '\right\}$

$= - \left\{\left({x}^{-} 2\right) {e}^{x} + {e}^{x} \left(- 2 {x}^{-} 3\right)\right\}$

$= - {e}^{x} \left({x}^{-} 2 - 2 {x}^{-} 3\right)$

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

$= - e \left(1 - 2\right) = e$.

Hence, the Reqd. Slope$= e$.