Let \(f\) be a function, and \(x\) a point in the domain of \(f\text{.}\) Given a real number \(D\text{,}\) we say that \(f\) has derivative \(D\) at the point \(x\) if the following two conditions hold:
If \(S\) is an open interval containing \(D\text{,}\) then there is an open interval \(T\) containing \(x\) such that if \(t\in T\setminus \set{x}\text{,}\) and \(t\) is in the domain of \(f\text{,}\) then
\begin{equation*}
\frac{f(t) - f(x)}{t-x}\in S.
\end{equation*}
If \(f\) does indeed have a derivative at some points in its domain, we call the derivative of \(f\), denoted \(f'\text{,}\) the function given by sending such an \(x\) to the derivative \(D\) at the point \(x\text{.}\) We write \(f'(x) = D\) in that case. We also say that \(f\) is differentiable at such points.
Note that the definition of the derivative automatically excludes the kind of behavior we saw with continuous functions, where a function defined only at a single point was continuous.
If \(f\) is differentiable at \(x\) and \(c\in\R\text{,}\) show that the function \(cf\) also has a derivative at \(x\) and \((cf)'(x) = cf'(x)\text{.}\)
If \(f\) and \(g\) are differentiable at \(x\text{,}\) show that the function \(f+g\) also has a derivative at \(x\) and \((f+g)'(x) = f'(x) + g'(x)\text{.}\)
Let \(f'(x) = D_f\) and \(g'(x) = D_g\text{.}\) Why can we restrict ourselves to intervals of the form \(S = \left(D_f + D_g - \frac{1}{n}, D_f + D_g + \frac{1}{n}\right)\) for some \(n\in \N\text{?}\)
When it comes to products, things are more subtle, as shown in stating the familiar product rule. On a different note, our facts about products from previous chapters can be used to show that differentiable functions are continuous. The following examples show the converse of this fact isn’t true, and that even when \(f'\) does make sense, the domain of \(f'\) might be a proper subset of the domain of \(f\text{.}\)
Suppose a function \(f\) defined on the closed interval \([a,b]\) has a derivative at some \(x\in (a,b)\) and \(f(x) \ge f(t)\) for all \(t\in (a,b)\text{.}\) Show that \(f'(x) = 0\text{.}\)
Suppose that \(a \lt b\text{,}\) that \(f\) is differentiable on \((a,b)\text{,}\) continuous on \([a,b]\text{,}\) and that \(f(a) = f(b) = 0\text{.}\) Show that there exists a point \(c\in (a,b)\) such that \(f'(c) = 0\text{.}\)
Now consider a more general function where \(f(a) \ne f(b)\text{.}\) If one carefully considers the secant line considering the endpoints of the graph of this function and applies Problem 134, one can prove the following crucial result, the Mean Value Theorem.
