Explanation

(a)
Let $y = x^{x}$
On differentiating w.r.t. $x$ , we get
$\frac{dy}{dt} = x^{x}(1 + \log x)$
For increasing function,
$\frac{dy}{dx} > 0$
$\Rightarrow \ \ x^{x}\left( 1 + \log{x)} > 0 \right)\ \ \ \Rightarrow \ \ 1 + \log x > 0$
$\Rightarrow \ \ \ \log_{e}x > \log_{e}\frac{1}{e}\ \ \ \ \Rightarrow \ \ \ x > \frac{1}{e}$
Function is increasing when $x > \frac{1}{e}$