Exercise 5.1.1.
Prove that if the limit (5.1.1) exists, it is equal to
\begin{equation*}
\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}
\end{equation*}
Section 5.1 The derivative
Subsection 5.1.1 Definition of derivative
Let \(I\) be an open interval in \(\mathbb R\text{,}\) and let \(c\in I\text{.}\) We say that a function \(f: I \rightarrow {\mathbb R}\) is differentiable at \(c\) if the limit
\begin{equation}
\lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}\tag{5.1.1}
\end{equation}
exists. In this case, we denote the limit by \(f'(c)\text{,}\) and we say that \(f'(c)\) is the derivative of \(f\) at \(x=c\text{.}\)
So, provided the limit exists, we can write
\begin{equation*}
f'(c)=\lim_{h\rightarrow 0}\frac{f(c+h)-f(c)}{h}=\lim_{x\rightarrow c}\frac{f(x)-f(c)}{x-c}.
\end{equation*}
If the limit does not exist, we say that \(f\) is not differentiable at \(x=c\text{,}\) or that \(f'(c)\) does not exist.
Exercise 5.1.2.
Show that \(f(x)=x^2\) is differentiable at every \(c\in {\mathbb R}\text{,}\) and \(f'(c)=2c\text{.}\)
Exercise 5.1.3.
Prove that \(f(x)=|x|\) is not differentiable at \(x=0\text{.}\) Find two sequences converging to \(0\) for which you can apply Theorem 4.1.4.
Hint.
Subsection 5.1.2 Basic results
Theorem 5.1.4.
Let \(f\) and \(g\) be both differentiable at \(x=c\text{,}\) and let \(a,b\) be real numbers. Then \(af+bg\) is differentiable at \(x=c\text{,}\) and
\begin{equation*}
(af+bg)'(c)=af'(c)+bg'(c)
\end{equation*}
Note that by suitable choice of \(a\) and \(b\) in the previous theorem, we obtain the familiar rules:
\begin{equation*}
(f\pm g)'(c)=f'(c)\pm g'(c), \hspace{3ex} (af)'(c)=af'(c).
\end{equation*}
Theorem 5.1.5.
Suppose \(f:I\longrightarrow {\mathbb R}\) is differentiable at \(x=c\text{.}\) Then \(f\) is continuous at \(x=c\text{.}\)
Hint.
Theorem 5.1.6. The product rule.
Suppose that \(f\) and \(g\) are both differentiable at \(x=c\text{.}\) Then \(fg\) is differentiable at \(x=c\text{,}\) and
\begin{equation*}
(fg)'(c)=f'(c)g(c)+f(c)g'(c)
\end{equation*}
Hint.
Theorem 5.1.7. The reciprocal rule.
Suppose that \(f\) is differentiable at \(x=c\text{,}\) and \(f(c)\neq 0\text{.}\) Then \(1/f\) is differentiable at \(x=c\text{,}\) and
\begin{equation*}
\left(\frac{1}{f}\right)'(c)=-\frac{f'(c)}{(f(c))^2}
\end{equation*}
Hint.
Theorem 5.1.8. The quotient rule.
Suppose \(f\) and \(g\) are both differentiable at \(x=c\text{,}\) and \(g(c)\neq 0\text{.}\) Then \(f/g\) is differentiable at \(x=c\) and
\begin{equation*}
\left(\frac{f}{g}\right)' = \frac{f'(c)g(c)-f(c)g'(c)}{(g(c))^2}
\end{equation*}
Hint.
Theorem 5.1.9. The chain rule.
Suppose that \(g\) is differentiable at \(x=c\) and \(f\) is differentiable at \(g(c)\text{.}\) Then \(f\circ g\) is differentiable at \(x=c\text{,}\) and
\begin{equation*}
(f\circ g)'(c) = f'(g(c))g'(c).
\end{equation*}
Hint.
\begin{equation*}
\frac{f(g(x))-f(g(c))}{x-c}=\frac{f(g(x))-f(g(c))}{g(x)-g(c)}\frac{g(x)-g(c)}{x-c}
\end{equation*}
and do the limit as \(x\rightarrow c\text{.}\) But \(g(x)-g(c)\) could be zero, so we cannot do that. To fix the problem, define a function \(\phi(t)\) by
\begin{equation*}
\phi(t)=\begin{cases} \frac{f(t)-f(g(c))}{t-g(c)} \amp \mbox{ if } t\neq g(c)\\
f'(g(c)) \amp \mbox{ if } t=g(c)\end{cases}.
\end{equation*}
Proof Define \(\phi\) as in the hint. Note that by definition of the derivative of \(f\) at \(g(c)\text{,}\)
\begin{equation*}
\lim_{t\rightarrow g(c)}\phi(t) = \lim_{t\rightarrow g(c)}\frac{f(t)-f(g(c))}{t-g(c)}=f'(g(c))=\phi(g(c)).
\end{equation*}
So \(\phi\) is continuous at \(g(c)\text{.}\) Now check that
\begin{equation*}
\frac{f(g(x)) - f(g(c))}{x-c}=\phi(g(x))\frac{g(x)-g(c)}{x-c}.
\end{equation*}
This is clearly true if \(g(x)\neq g(c)\text{,}\) and it is also true if \(g(x)=g(c)\) because in that case both sides are \(0\text{.}\) Taking the limit as \(x\rightarrow c\) and using continuity of \(g\) and \(\phi\text{,}\) we find our result.
Theorem 5.1.10.
[The derivative of the inverse function]
