Determine the values of `x` for which `f(x)=x^x , x >0` is increasing or decreasing.

Explanation

`frac{dy}{dx}=x^x(1+log x)`

For incraesing function, `frac{dy}{dx}>0`

`=>x^x(1+logx)>0`

`1+logx>0`

`log_e x>log_efrac{1}{e}`

`x>1/e`

Therefore, the function is increasing, when `x>1/e`