In Binomials are the answer!, we proved that the derivative of \(x^n\) is \(nx^{n-1}\) for any positive integer \(n\).

Using exponentials and/or logarithms, can you now prove that the derivative of \(x^a\) (where \(x>0\) and \(a\) is any real number) is \(ax^{a-1}\)?