Show that `f(x)=e^(1//x),\ \ x!=0` is a decreasing function for all `x!=0` .

Explanation

`f(x)=e^(1//x),\ \ x!=0`
`f'(x)=e^(1/e)-1/(x^2)`

`=frac{-e^(1/x)}{x^2}`

for the function be decreasing or increasing `f
′
(x)` should be less than or more than equal to `0`
since the function is decreasing
`e^(1/x)>0 ∀x
!
...