\begin{gather*} \left(x^{x}\right)'=\left(e^{\ln x^{x}}\right)'=\left(e^{x\cdot\ln x}\right)'=\left(e^{x\cdot\ln x}\right)\cdot\left(\ln x+1\right)=x^{x}(\ln x+1)\end{gather*} Innen elég a kitevot deriválni: $\left(x\cdot\ln x\right)'=1\cdot\ln x+x\cdot\dfrac{1}{x}=\ln x+1$