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.

Just do the same thing:  $$\log \left( \frac{dy}{dx} \right) = x \log x + \log (\log x + 1),$$ so $$\frac{1}{\frac{dy}{dx}} \frac{d^2 y}{dx^2} = (\log x + 1) + \frac{1}{\log x + 1} \cdot \frac{1}{x},$$ hence $$\frac{d^2 y}{dx^2} = \frac{dy}{dx} \left( \log x + 1 + \frac{1}{x(\log x + 1)} \right),$$ and substitute the expression you obtained for the first derivative.
Alternatively, $$\frac{dy}{dx} = x^x (\log x + 1)$$ implies $$\frac{d^2 y}{dx^2} = \frac{d}{dx}\left[x^x \right] (\log x + 1) + x^x \cdot \frac{1}{x}$$ by the product rule, and as before, substitute the derivative of $x^x$ that you found earlier.
