Find the derivative of $y=x^x$. Find the derivative of $y=x^x$.
My Attempt:
$$y=x^x$$
Taking $\textrm {ln}$ on both sides, we get:
$$\textrm {ln} y= \textrm {ln} x^x$$
$$\textrm {ln} y = x \textrm {ln} x$$
How do I procees further?

your start is very good: $$\frac{1}{y}y'=\ln(x)+x\frac{1}{x}$$
