What is the derivative of $x^{x^x}$?
$y=x^{x^x}$
We need to find derivative of $y $with respect to $x.$ So solve it using log-
$ln\,y = x^x\,\ln\,x$
Now, use the product formula
$\dfrac{d}{dx}(f(x)g(x))=f(x)\dfrac{d}{dx}g(x)+g(x )\dfrac{d}{dx}f(x)$
Here,
$f(x)=x^x$
$g(x)=\ln\,x$
Substitute values in formula-
$\dfrac{1}{y}\dfrac{dy}{dx}=(x^x)\dfrac{d}{dx} \ln\,x+\ln\,x\dfrac{d}{dx}x^x $
$\dfrac{1}{y}\dfrac{dy}{dx}=\dfrac{1}{x}x^x+x^x( \ln \,x+1)\ln\,x $
$\dfrac{dy}{dx}=y(x^{x-1}+x^x(\ln\,x+1)\ln\,x) $
$\dfrac{dy}{dx}=x^{x^x}(x^{x-1}+x^x(\ln\,x+1)\ln\,x) $