Finding the second derivative of $x^x$ 
Find the second derivative $d^2y/dx^2$ when $y=x^x\:(x>0)$. 

$$y=x^x,\:\:(x\gt0)$$
\begin{align}
\log y&=x\log x \\
\rm{Differentiating}&\:{\rm{with\:respect\:to\:}}x
\end{align}
\begin{align}
\frac{1}{y}\frac{dy}{dx}&=1\cdot(\log x+1)+x\cdot\frac{1}{x} \\[0.8ex]
\frac{dy}{dx}&=x^x(\log x+1)
\end{align}
I found the first derivative, and now I want to know how to find the second derivative of this function.

$y = x^x$
As shown, $\frac{dy}{dx} = x^x(\log x + 1)$
Differentiating, again w.r.t. $x$, we get,
$\frac {d^2y}{dx^2} = \frac{d}{dx}{x^x (\log x + 1)}$
Using given information and differentiating by parts, we get,
$$\frac {d^2y}{dx^2} = (\log x + 1)\frac{d}{dx}x^x + x^x \frac{d}{dx}(\log x + 1)$$
$$\frac {d^2y}{dx^2} = (\log x + 1).x^x(\log x + 1) + x^x(\frac {1}{x} + 0)$$
$$\frac {d^2y}{dx^2} = x^x(\log x + 1)^2 + x^{x-1}$$
Required answer.
