Fun with derivatives

$y=x^x$ can be written as $\log_x y = x$

$\Rightarrow$ $\frac{\log_e y}{\log_e x}=x$

$\Rightarrow$ $\log y = x \times log x$

differentiating both sides w.r.t $x$ , we get :

$\frac{1}{y} \times \frac{dy}{dx} = x \times \frac{1}{x} + \log x$

solving this for $\frac{dy}{dx}$ , we get :

$\frac{dy}{dx} = y(1 + \log x)$ -------------------------(1)

given, $y=x^x$ -------------------------------------------(2)

hence, by substituting (2) in (1),

$\boxed{x^x(1 + \log x) }$

y = e^(x Ln x)

Let u = x Ln x, y = e^u

d u/ d x = Ln x + x (1/ x) = Ln x + 1

d y/ d u = e^u

d y/ d x = (d y/ d u)(d u/ d x) = e^(x Ln x)(Ln x + 1)

d y / d x = x^x (1 + Ln x)