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

##### 1 Answer
May 11, 2016

Slope of $f \left(x\right)$ at $x = 1$ is $0$

#### Explanation:

$f \left(x\right) = - x \cdot {e}^{x} / {x}^{2}$ $= - {e}^{x} / x$ = $- {e}^{x} \cdot {x}^{-} 1$

$f ' \left(x\right) = - \left({e}^{x} \cdot \left(- {x}^{-} 2\right) + {x}^{-} 1 \cdot {e}^{x}\right)$ (Product, Exponential and Power Rules)

The slope of $f \left(x\right)$ at $x = 1$ is given by $f ' \left(1\right)$

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

$f ' \left(1\right) = - \left(- e + e\right) = 0$

Hence: Slope of $f \left(x\right)$ at $x = 1$ is $0$