Exercise 5.6.1.
- Show that \(f(x)=\sqrt[3]{x}\) is not differentiable at \(x=0\text{.}\)
- Explain why asking if \(f(x)=\sqrt{x}\) is differentiable at \(x=0\) is not a well-defined question.
Section 5.6 Exercises
Exercise 5.6.2.
Prove Theorem 5.1.4
Exercise 5.6.3.
Let \(I\) be an open interval, \(c\in I\text{,}\) and \(f: I\rightarrow {\mathbb R}\text{.}\)
- Prove that \(f\) is differentiable at \(x=c\) if and only if there is a continuous function \(\phi\) such that \(f(x)-f(c)=\phi(x) (x-c)\) holds for all \(x\in I\text{.}\)
- Prove that if \(f\) is differentiable at \(x=c\text{,}\) and \(\phi\) is as in part a., then \(f'(c)=\phi(c)\)
Exercise 5.6.4.
Use Exercise 5.6.3 to give a different (and more elegant) proof of the chain rule. Write down the condition for differentiability given by part a. of Exercise 5.6.3 for \(f\) at \(g(c)\text{,}\) and for \(g\) at \(c\text{.}\)
Hint.
Proof Since \(g\) is differentiable at \(c\text{,}\) there is a continuous function \(\phi\) such that
\begin{equation}
g(x)-g(c)=\phi(x)(x-c),\tag{5.6.1}
\end{equation}
and \(\phi(c)=g'(c)\text{.}\) Since \(f\) is differentiable at \(g(c)\text{,}\) there is a continuous function \(\psi\) such that
\begin{equation}
f(g(x))-f(g(c))=\psi(x)(g(x)-g(c)),\tag{5.6.2}
\end{equation}
\begin{equation*}
(f\circ g)(x)-(f\circ g)(c)=f(g(x))-f(g(c))=\psi(x)\phi(x)(x-c).
\end{equation*}
Since \(\psi \phi\) is continuous, from Exercise 5.6.3 part a. we conclude that \(f\circ g\) is differentiable at \(x=c\text{,}\) and then from part b. we get that \((f\circ g)'(c) = \psi(c)\phi(c)= f'(g(c))g'(c)\text{.}\)
