Derivatives of Monomials

We have seen [Derivative] that the derivative for a function \(f\) at some point \(x_0\) is given formally by \[ f'(x_0)=\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}. \] However, this does not do us much good - how can we find this in practice? In this section, we'll compute some derivatives of some common functions, with the plan of building a library of things from which we can construct more complicated derivatives.

The simplest function whose derivatives we can compute is a constant function \(f\) such that \(f(x)=K\). In that case it is immediate that \[f'(x_0)=\lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}= \lim_{x\to x_0}\frac{K-K}{x-x_0} = 0. \] This is no surprise, inasmuch as it is obvious that for a constant function we always have \(f(x)=f(x_0)\) for every point \(x_0\) in the domain of \(f\). In other words, using the formulation from Section 1: \[f(x) = f(x_0)+0(x-x_0) + 0(x-x_0)^2.\]

The next obvious thing to consider is the function given by \(f(x)=x\). In this case \begin{align} f'(x_0) &= \lim_{x\to x_0}\frac{f(x)-f(x_0)}{x-x_0}\\ &=\lim_{x\to x_0}\frac{x-x_0}{x-x_0}\\ &= 1. \end{align} Again this is no surprise, since \begin{align} f(x) &= f(x_0)+f'(x_0)(x-x_0)+R(x)(x-x_0)^2\\ &= x_0 + (x-x_0). \end{align}

Let's keep going to consider the function given by \(f(x)=x^2\). This time we have \begin{align} f'(x_0) &= \lim_{x\to x_0}\frac{x^2-x_0^2}{x-x_0}\\ &= \lim_{x\to x_0}\frac{(x-x_0)(x+x_0)}{x-x_0}\\ &= \lim_{x\to x_0}{x+x_0}\\ &= 2x_0. \end{align} This seems a bit less obvious. On the other hand, we might notice a pattern here. As we increase the powers on \(x\) we will always be able to factor out a term that looks like \((x-x_0)\). That will always cancel with the term in the denominator. We could do a couple more powers of \(x\) to verify it, but let's just go to a hypothesis about the derivative of \(x^n\) for non-negative \(n\). We think that whenever \(f(x)=x^n\), then \(f'(x_0)=nx_0^{n-1}\). Let's see whether we can establish that result. We can write \begin{align} f'(x_0) &= \lim_{x\to x_0}\frac{x^n-x_0^n}{x-x_0}\\ &= \lim_{x\to x_0}\frac{(x-x_0)(x^{n-1}+x^{n-2}x_0+x^{n-3}x_0^2+\cdots+x_0^{n-1})}{x-x_0}\\ &= \lim_{x\to x_0}(x^{n-1}+x^{n-2}x_0+x^{n-3}x_0^2+\cdots+x_0^{n-1})\\ &= nx_0^{n-1}. \end{align}

This is such a nice result. Does it work for negative powers of \(x\)? Let \(f(x)=x^{-n}\) for some positive integer \(n\). Then \begin{align} f'(x_0) &= \lim_{x\to x_0} \frac{\frac1{x^n}-\frac1{x_0^n}}{x-x_0}\\ &= \lim_{x\to x_0} \frac{\frac{x_0^n}{x_0^nx^n}-\frac{x^n}{x^nx_0^n}}{x-x_0}\\ &= \lim_{x\to x_0} \frac{x_0^n-x^n}{(x^nx_0^n)(x-x_0)}\\ &= \lim_{x\to x_0} \frac{(x_0-x)(x^{n-1}+x^{n-2}x_0+x^{n-3}x_0^2+\cdots+x_0^{n-1})}{(x^nx_0^n)(x-x_0)}\\ &= -\lim_{x\to x_0} \left( \frac{x^{n-1}}{x^nx_0^n} + \frac{x^{n-2}x_0}{x^nx_0^n} + \cdots + \frac{x_0^{n-1}}{x^nx_0^n} \right)\\ &= -\frac{n}{x_0^{n+1}}\\ &= -n x_0^{-n-1},\\ \end{align} so long as \(x_0\ne 0\). Thus, we get the following remarkable result.

Theorem: [MonomialDerivative] For any integer \(n\) the derivative of \(x^n\) is \(nx^{n-1}\), wherever that is defined.