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 $x=1$.


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 $x=1$. 
