Building blocks

## Solution

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 $ax^{a-1}$?

If we write $x=e^{\ln x}$, then $x^a$ becomes $e^{a\ln x}$. If we write $u=a\ln x$, then we can differentiate this using the chain rule (in the form $\frac{dy}{dx}=\frac{dy}{du}\times\frac{du\vphantom{y}}{dx}$, with $y=e^u$): \begin{align*} \frac{d}{dx}(e^{a\ln x})&=\frac{d}{du}(e^u)\times \frac{d}{dx}(a\ln x)\\ &=e^u\times\frac{a}{x}\\ &=e^{a\ln x}\times\frac{a}{x}\\ &=x^a\times\frac{a}{x}\\ &=ax^{a-1} \end{align*}

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.