\[f(x)=x^x\]
us a super exponential and can be differentiated by writing \[f(x)=e^{ln(x^x}=e^{x lnx}\]
and using a combination of The Chain Rule and Differentiation - The Product Rule.Using the product rule to differentiate
\[xlnx\]
gives \[x \times \frac{1}{x} +1 \times lnx=1+lnx\]
.Using the chain rule to differentiate
\[e^{x lnx}\]
gives \[(1+lnx)e^{xlnx}=(1+lnx)x^x\]
.