Derivative

Let \(f\) be a function. For \(x_0\), and \(x\), we can define a function on \(h\) as \([f(x)-f(x_0)]/[x-x_0]\). That measures the average change in \(f\) between \(x_0\) and \(x\). It is not well-defined, of course, at \(x_0\) itself. However, that does not stop us from asking about what limit it has at \(x_0\). There might, of course, not be a limit. If there is one, however, we call that limit “the derivative of \(f\) at \(x_0\)”, and write \(f'(x_0)\).

We note that if \(f'(x_0)\) exists, then \(f\) must be continuous at \(x_0\).

From the definition, and the fact that limits of sums are sums of limits, we see that \((f+g)'=f'+g'\). With a bit of algebra, it is easy to see that \((fg)'=f'g+fg'\). If \(f\) and \(g\) are functions, we define \((f\circ g)(x)=f(g(x))\). With this definition, and the kind of \(\epsilon\)/\(\delta\) games we are used to, it is straightforward to show that (fcirc g)’(x_0)=f’(g(x_0))g’(x_0).

Assume \(f'(x_0)>0\). Let \(\epsilon=f'(x_0)/2\). Then by definition of limit, there is a \(\delta\), \(0<|x-x_0|<\delta\) implies that \(|[f(x)-f(x_0)]/[x-x_0]-f'(x_0)|f'(x_0)-\epsilon=f'(x_0)/2>0\). If \(0<x-x_0f(x_0)\), and if \(0<x_0-x<\delta\) then \(f(x)<f(x_0)\). Therefore, on any segment that contains \(x_0\) (except as an endpoint) \(f(x_0)\) is not a minimum or a maximum. For analogous reasons, if \(f'(x_0)<0\), it is neither a minimum or a maximum. In other words, if \(f'(x_0)\) is a minimum or a maximum, even if only on a short segment that includes \(x_0\) not as an endpoint, then \(f'(x_0)=0\).