To verify that

holds for all real $a$ and positive $x$.

Solution. We recall that $x^{a}=\exp(a\log x)$, and then we differentiate

To find ${{d}\over{dx}}(x^{x})$. (Here $x$ appears both as the variable and as the power.)

Solution. Let $y=x^{x}$ and $u=\log y=x\log x$; then $y=e^{u}$ and so by the chain rule