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 \(ax^{a-1}\)?
as we wanted.
This idea of writing \(x^a\) as \(e^{a\ln x}\) is actually the simplest and most useful way to define arbitrary powers. If we tried to define powers in a different way, it would turn out to be quite difficult to prove that the derivative of \(x^{\pi}\) is \(\pi x^{\pi-1}\), for example.
You may also have thought of proving this result using the general binomial theorem using a similar approach to that in Binomials are the answer!, as follows.
The derivative of \(x^a\) is the limit of \(\dfrac{(x+h)^a-x^a}{h}\) as \(h\to0\). Then using the binomial theorem, we have \[\begin{align*} (x+h)^a &= x^a\left(1+\frac{h}{x}\right)^a\\ &= x^a\left(1+a\cdot\frac{h}{x}+\frac{a(a-1)}{2}\cdot\left(\frac{h}{a}\right)^2+\cdots\right). \end{align*}\]Then subtracting \(x^a\) and dividing by \(h\) gives \[\dfrac{(x+h)^a-x^a}{h} = x^a\left(a\cdot\frac{1}{x}+\frac{a(a-1)}{2}\cdot\frac{h}{a^2}+\cdots\right)\] As \(h\to0\), the right hand side tends to \(x^a\cdot\frac{a}{x}=ax^{a-1}\), as required.
It turns out that there is a subtle difficulty with this approach. It uses the general binomial theorem for \((1+x)^a\), but how do we prove that? The usual way to do so is to use Taylor’s theorem, but that requires us to already know the derivative of \(x^a\)…. So this would be a circular argument! But now that we can differentiate \(x^a\) without reference to the binomial theorem, we can use this to prove the general binomial theorem, and all works nicely.
