Differentiate x^x

With the methods we know at A Level we cannot current differentiate x^x in its current form. Therefore let y = x^xTo turn it into a form we can differentiate we take the natural log of both sides. This gives ln(y) = ln(x^x). Using the log rule (loga^b = bloga) we can then say ln(y) = xln(x). We can then differentiate implicitly to form a differential equation 1/y x dy/dx = ln(x) + 1. To find dy/dx we then simply multiply through by y to give...dy/dx = y(lnx + 1) = x^x(lnx + 1)