Problem - 4594
Find the derivative of $x^x$.
Let $f(x)=x^x$, then
$$f(x)=e^{x\ln{x}}\implies f'(x)=e^{x\ln x}\left(\ln{x} + x\cdot\frac{1}{x}\right)=x^x(\ln{x}+1)$$
Find the derivative of $x^x$.
Let $f(x)=x^x$, then
$$f(x)=e^{x\ln{x}}\implies f'(x)=e^{x\ln x}\left(\ln{x} + x\cdot\frac{1}{x}\right)=x^x(\ln{x}+1)$$