Chapter 4 Differentiation
It's time for calculus.
Definition 119.
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 \(f\) 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.
Example 120.
A function defined only on \(\Z\) cannot have a derivative.
Example 121.
Find a formula for the derivative of \(f(x) = 3\) on \(\R\text{,}\) and prove your formula is correct.
Problem 122.
Find a formula for the derivative of \(f(x) = 2x - 5\) on \(\R\text{,}\) and prove your formula is correct.
Problem 123.
Find the derivative of \(h(x) = x^2 - x +1\) at \(x=2\text{.}\)
Problem 124.
Find the derivative of \(h(x) = x^2 + ax + b\) for any \(a,b\in \R\text{.}\)
Problem 125.
If \(f\) is differentiable at \(x\) and \(c\in\R\text{,}\) show that the function \(cf\) also has a derivative at \(x\) and \((cf)'(x) = cf'(x)\text{.}\)
Problem 126.
If \(f\) and \(g\) are differentiable at \(x\text{,}\) show that the function \(f+g\) also has a derivative at \(x\) and \((f+g)'(x) = f'(x) + g'(x)\text{.}\)
Hint.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 \(f\text{.}\)
Problem 127.
Let \(f(x) = |x|\text{.}\) Prove that \(f\) is differentiable everywhere except at \(x=0\text{.}\)
Problem 128.
Let \(f(x) = x\) if \(x\in\Q\text{,}\) and \(f(x) = 0\) otherwise. Show that \(f\) is continuous at \(x=0\text{,}\) but not differentiable.
Example 129.
There is a function like this which is differentiable at \(x=0\text{!}\)
It's useful to spend just a little time thinking about how one would prove the usual facts about derivatives used in a calculus course.
Problem 130.
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 \(f'(x) = 0\text{.}\)
Hint.Consider \(f'(x) = D\) and show that we cannot have \(D\lt 0\) or \(D\gt 0\text{.}\)
So if there is an extreme value of a differentiable function, it occurs at a critical point. This leads us to (a variant of) Rolle's Theorem.
Problem 131.
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 131, one can prove the following crucial result, the Mean Value Theorem.
Problem 132.
Suppose \(f\) is a continuous function on \([a,b]\) and differentiable on \((a,b)\text{.}\) Then there is a point \(c\in (a,b)\) such that
\begin{equation*}
f'(c) = \frac{f(b) - f(a)}{b-a}.
\end{equation*}
Hint.Let \(g(x) = \frac{f(b) - f(a)}{b-a} (x-a) + f(a)\) and consider the function \(h(x) = f(x) - g(x)\text{.}\)
Among many other wonderful things this theorem provides, we have the following standard result.
Problem 133.
If the derivative of a function is uniformly zero on a closed interval, then the function is constant on that interval.
