Building blocks

Powerful derivatives Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Problem

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 $ax^{a-1}$?