**Maintainer:**admin

Today we finally discuss differentiability and the definition of the derivative. Recall from introduction to calculus that

$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} $$

The equivalent formulation that we'll use is:

$$ f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c} $$

**Definition:** Let $I \subseteq \mathbb{R}$ be an interval, $f : I \to \mathbb{R}$ and $c \in I$. We say $L \in \mathbb{R}$ is the derivative of $f$ at $c$ if for a given $\epsilon > 0$, there exists a $\delta > 0$ such that if $0 < |x-c| < \delta$ then:

$$
\left| \frac{f(x)-f(c)}{x-c} - L \right| < \epsilon
$$

In this case we say $f$ is *differentiable* at $c$ and denote $L = f'(c)$.

As you must remember, the derivative of $f$ is only defined at $c$ if $f$ is continuous at $c$. Thus it's a pretty simple proposition that:

**Theorem:** Let $I$ be an interval, $f : I \to \mathbb{R}$ be a function on $I$, and $c \in I$. If $f$ is differentiable at $c$ then $f$ is continuous at $c$.

**Proof:** Let $\epsilon > 0$ be given - WLOG, take $\epsilon < 1$. As $f$ is differentiable, there exists some $L \in \mathbb{R}$ and $\delta_1 > 0$ such that if $0 < |x-c| < \delta_1$, then $\left|\frac{f(x)-f(c)}{x-c} - L\right| < \epsilon$. Let $\delta = \inf\{\delta_1, \frac{\epsilon}{1+|L|}\}$. Then if $0 < |x-c| < \delta$, it follows that

$$ |f(x) - f(c)| = \left|\frac{f(x)-f(c)}{x-c}(x-c)\right| \\ = \left|\left[\frac{f(x)-f(c)}{x-c}-L+L\right](x-c)\right| \\ = \left|\left[\frac{f(x)-f(c)}{x-c}-L\right](x-c) + L(x-c)\right| \\ \leq \left|\frac{f(x)-f(c)}{x-c}-L\right|\left|x-c\right| + |L|\left|x-c\right| \\ < |x-c| + |L||x-c| = (1 + |L|)|x-c| \\ < \frac{\epsilon (1 + |L|)}{1+|L|} = \epsilon \qquad \blacksquare $$

Note that this is only a one-way implication - continuity does not guarantee differentiability. Consider $f(x) = |x|$, for example - continuous but not differentiable at $x=0$.

Now we come to the theorems we know and love:

**Theorem:** Let $I \subseteq \mathbb{R}$ be an interval, $c \in I$, $f, g : I \to \mathbb{R}$ be functions that are differentiable at $c \in I$. Then:

(i) if $k \in \mathbb{R}$, $kf$ is differentiable at $c$ and $(kf)'(c) = kf'(c)$

(ii) $f + g$ is differentiable at $c$ and $(f + g)'(c) = f'(c) + g'(c)$

(iii) $fg$ is differentiable at $c$ and $(fg)'(c) = f(c)g'(c) + g'(c)f(c)$

(iv) if $g(c) \neq 0$ then $\frac{f}{g}$ is differentiable at $c$ and

$$ \left(\frac{f}{g}\right)'(c) = \frac{g(c)f'(c) - f(c)g'(c)}{g^2(c)} $$

The proofs of these statements can be found in the book, Section 6.1. (i) and (ii) in particular are almost trivial.