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\begin{center} \textbf{The Derivative.}
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Reminder: If $D \subset R$ then $Int(D)$ denotes the set to which $x$ belongs if and only if there is an open set $U$ so that $x \in U \subset D$.

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Definition.  Suppose that $f: D \rightarrow \mathbb{R}$ is a function and $p \in Int(D)$.  Then $f$ is said to be \textit{differentiable} at $p$ means that there exists a number $m$ so that
$$\lim_{h \rightarrow 0} \frac{f(p+h) - f(p)}{h} = m.$$
The number $m$ is called the derivative of $f$ at the point $p$ and $m$ is denoted by $f'(p)$.

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Theorem 1.1.  If the function $f$ is differentiable at the point $p$ then it is continuous at the point $p$.

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Question.  Is the converse of Theorem 1.1 true?

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Theorem 1.2 A.  The function  $f: D \rightarrow \mathbb{R}$ is differentiable at the point $p$ if and only if:
$$\lim_{x \rightarrow p} \frac{f(p) - f(x)}{p-x} = f'(p).$$

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Theorem 1.2 B [The geometric picture of the derivative].   Suppose that $f: D \rightarrow \mathbb{R}$ is a function and $p \in Int(D)$.  Then $f$ is differentiable at $p$ if and only it there exists a number $m$ so that if $U$ is an open set containing $f(p)$ and $\alpha$ is a line containing $(p,f(p))$ with slope less than $m$ and $\beta$ is a line containing $(p,f(p))$ with slope bigger than $m$ then there is an open set $V$ containing $p$ so that if $x \in V-\{p\}$ then $f(x)$ lies between $\alpha$ and $\beta$.  Furthermore $m = f'(p)$.



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Theorems 1.3.  Suppose that $f$ and $g$ are functions whose domains contain $p$ in their interiors, $c$ is the constant function and all the quantities in the following formulas are defined. Then:

\begin{eqnarray}
c'(p) & = & 0 \\
(cf)'(p) & = & c f'(p) \\
(f+g)'(p) & = & f'(p) + g'(p) \\
(fg)'(p) & = & f'(p) g(p) + f(p) g'(p) \\
\Big{(}\frac 1f \Big{)}'(p) & = & - \frac{f'(p)}{(f(p))^2} \\
(f \circ g)'(p) & = & f'(g(p)) g'(p)
\end{eqnarray}

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Theorem 1.4.  Suppose that $r \in R$ and $f$ is defined by $f(x) = x^r$.  Then $f'(x) = rx^{r-1}$.

[Hint: Prove versions of the theorem for the following cases:

\qquad (i.) $r \in \mathbb{N}$;

\qquad (ii.) $r \in \mathbb{Z}$;

\qquad (iii.) $\frac 1r \in \mathbb{Z}, r \ne 0$;

\qquad (iv.) $r \in \mathbb{Q}$;

\qquad (v.) $r \notin \mathbb{Q}$.]

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Theorem 1.5 (Rolle's theorem, Relative Max/Min theorem).  Suppose that $f$ is a function that is differentiable at $p$ and that there is an open set $V$ containing $p$ so that $f(x) \le f(p)$ for all $x \in V$.  Then $f'(p)=0$.

Lemma to Theorem 1.5.  If $f: [a,b] \rightarrow \mathbb{R}$ is a function, $a < p < b$ and $f'(p) \ne 0$, then if $\delta > 0$ there is a point $q \in (p-\delta, p+\delta)$ so that $f(q) > f(p)$.

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Theorem 1.5B (Rolle's theorem, Relative Max/Min theorem).  Suppose that $f$ is a function that is differentiable at $p$ and that there is an open set $V$ containing $p$ so that $f(x) \ge f(p)$ for all $x \in V$.  Then $f'(p)=0$.

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Corollary to Theorem 1.5.  Suppose that $f: [a,b] \rightarrow \mathbb{R}$ is a continuous function that is differentiable at each point of $(a,b)$ and $f$ satisfies the hypothesis of Theorem 1.5.  Then there is a point $p$ between $a$ and $b$ so that $f'(p) = 0$.

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Question.  If $f$ is differentiable at each point of a segment $(a, b)$ then must $f$ be continuous on $(a,b)$.

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Theorem 1.6 (Mean value theorem).  Suppose that $f$ is continuous on the interval $[a,b]$ and is differentiable at each point of the segment $(a,b)$.  Then there is a point $c$ between $a$ and $b$ so that
$$f'(c) = \frac{f(b) - f(a)}{b-a}.$$

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Theorem 1.7 (Intermediate value theorem for derivatives).  Suppose that $f$ is differentiable on the interval $[a,b]$ and $m$ is between $f'(a)$ and $f'(b)$.  Then there is a point $z$ between $a$ and $b$ so that $f'(z) = m$.

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