How would you differentiate x^x?

To start off with, we have the expression y = x^x and we want to differentiate it. A clever way to do this would be to first remember implicit differentiation and start by taking the natural logarithm of both the sides giving us ln y = ln x^x. Now we want to remember the power property of logarithms which says that ln(a)^b = b * ln(a). Hence, we can write the expression as ln y = x * ln x. Now we diffenrentiate both sides and obtain 1/y * dy/dx = 1 * ln x + x * 1/x. This gets us (dy/dx)/y = ln x + 1. Now taking y to the other side we are left with dy/dx = (ln x + 1)y which is equivalent to dy/dx = x^x(ln x + 1).