If $y=e^{x^x},\text{ then }\cdot \frac{dy}{dx}=$

Now, $y=e^{x^x}$ 

Taking logarithms with base e, we get

${\log }_{e}y={\log }_e{e}^{x^x}$

${\log }_{e}y=x^x\cdot {\log }_{e}e=x^x,\left\{\because {\log }_{e}e=1\right\}$

Again taking logarithms with base e, we get,

${\log }_e\left({\log }_{e}y\right)={\log }_e{x}^x\text{ or }{\log }_e\left({\log }_{e}y\right)=x{\log }_{e}x$

Differentiating both sides with respect to x, we get

$\frac{1}{{\log }_{e}y}\cdot \frac{1}{y}\cdot \frac{dy}{dx}=1\cdot {\log }_{e}x+x\cdot \frac{1}{x}$ or $\frac{dy}{dx}=y{\log }_{e}y\cdot (\log x+1)$

$=e^{x^x}\cdot x^x.\left({\log }_{e}x+1\right).=y\cdot x^x\left(1+{\log }_{e}x\right)$