### Chain Rule & Integration by Substitution

Building blocks

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 $ax^{a-1}$?