Derivative of x^x

$f(x)=x^x=e^{x\ln(x)}$ $\therefore f'(x)=(\ln(x)+1)e^{x\ln(x)}=(\ln(x)+1)x^x$ $\implies f'(1)=(\ln(1)+1)\times1^1=\boxed{1}$

Let's say, y=x^x

thus, lny =ln(x^x)

ln y = x.lnx

By differentiating,

(1/y)*dy/dx = lnx + 1 (chain rule and uv for second part)

Thus, dy/dx= y*((lnx)+1)

Because y=x^x,

The result is (x^x)*((lnx)+1)

Now plug in values

Derivative of 3x^2

$(x^x)'=x^{x-1}+x^x log(x) |(x=1) = \boxed{1}$