# 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.