## anonymous 5 years ago what is the derivative of x^x?

Let $$y = x^x$$ $$\implies ln(y) = x*ln(x)$$ $$\implies \frac{d}{dx} ln (y) = \frac{d}{dx}[x * lnx]$$ $$\implies \frac{d}{dy} ln(y) * \frac{dy}{dx} = x*\frac{d}{dx}ln(x) + ln(x) * \frac{d}{dx}x$$ (chain rule) ( product rule ) $$\implies \frac{1}{y}* \frac{dy}{dx} = \frac{x}{x} + ln(x)$$ $$\implies \frac{dy}{dx} = y(1+ln(x))$$ $$=x^x(1+ln(x))$$