Use Logarithmic Differentiation to find $\frac{d}{dx} (x^{{x}^{x}})$ at $x=1$ How do  do this?

Use Logarithmic Differentiation to find $\frac{d}{dx} (x^{{x}^{x}})$ at $x=1$.

 A: Hint: $x^{x}=y$ . take a log of both side  you get $x$log$x$=log$y$. take derivative. then  you will get
$\dfrac{d}{dx} (x^{x})$$ = y'= x^{x}($log$x+1$) . Now write $x^{x^{x}}=y$ take log of both side . then $x^{x}$log$x=$log$y$. Now take derivative. I hope you can do the rest.
A: You can use the chain rule, amongst a variety of methods.  
Expressing $x^{x^x}$ as $e^{x^x\ln x}$, we can set $f(x)=x^x\ln x$.
Thus the differentiation is as follows:-
$$\frac{d}{dx}x^{x^x}=\frac{d}{dx}e^{x^x\ln x}=\frac{d}{dx}e^{\color{red}{f(x)}}\\=\frac{d(e^\color{red}{f(x)})}{d\color{red}{f(x)}}\frac{d(\color{red}{f(x)})}{dx}\\=\frac{d(e^\color{red}{f(x)})}{d\color{red}{f(x)}}\frac{d(x^x\ln x)}{dx}\\=e^{\color{red}{f(x)}}\left(x^x\frac{d(\ln x)}{dx}+\ln x\frac{d(x^x)}{dx}\right)\\=x^{x^x}\left(x^{x-1}+\ln x\left(\frac{d}{dx}(x\ln x)\right)e^{x\ln x}\right)\\=x^{x^x}\left(x^{x-1}+\ln x\left((1+\ln x)e^{x\ln x}\right)\right)\\=x^{x^x}(x^{x-1}+x^x(\ln x)(1+\ln x))$$
Having obtained the derivative, all you need to do is to set $x$ to $1$ to find the derivative at $x=1$. 
