Lecture 28 - The Derivative

We start our discussion with a definition:

derivative

Let $f : I \to R$ where $I$ is an interval (open, closed, half-open/closed), which is definitely connected. Then $f$ is differentiable at $c \in I$ if:
$lim_{x \to c} \frac{f (x) - f (c)}{x - c}$
exists. (see Functional Limit for details, and highlighting Existence of Functional Limits for showing this).

Note that:

If $f$ is differentiable at $c$ , then we define $f^{'} (c)$ to refer the value of the limit:
$lim_{x \to c} \frac{f (x) - f (c)}{x - c} = f^{'} (c)$
specifically we call this the derivative of $f$ at $c$ .
Similar to One-Sided Limits, if $c \in I$ is an endpoint of $I$ (ex: a closed interval), then this limit coincides with a one-sided limit (which we refer as a one-sided derivative):
$lim_{x \to c^{-}} \frac{f (x) - f (c)}{x - c}, lim_{x \to c^{+}} \frac{f (x) - f (c)}{x - c}$
If $f$ is differentiable at every $c \in I$ then we say $f$ is differentiable on $I$ .
$\frac{f (x) - f (c)}{x - c}$ is the slope of the secant line. Taking its limit gets the tangent line at a point.

Theorem

Let $f : I \to R$ where $I$ is an interval. If $f$ is differentiable at $c \in I$ , then $f$ is continuous at $c$ .

We'll be using the Algebraic Limit Theorem for Functional Limits

Proof

For all $x \in I$ where $x \neq c$ then we have the following:
$\begin{aligned} f (x) - f (c) & = \frac{f (x) - f (c)}{x - c} \cdot (x - c) \\ lim_{x \to c} [f (x) - f (c)] & = lim_{x \to c} [\frac{f (x) - f (c)}{x - c} \cdot (x - c)] \\ lim_{x \to c} f (x) - f (c) & = (lim_{x \to c} \frac{f (x) - f (c)}{x - c}) (lim_{x \to c} (x - c)) \\ = f^{'} (c) \cdot 0 & ALT \\ lim_{x \to c} [f (x) - f (c)] & = 0 \end{aligned}$
Then if we want to compute the limit of $f (x)$ at $c$ can be calculated:
$\begin{aligned} lim_{x \to c} f (x) & = lim_{x \to c} (f (x) - f (c)) + f (c) \\ = lim_{x \to c} f (x) - f (c) + lim_{x \to c} f (c) \\ = 0 + f (c) \\ = f (c) \end{aligned}$
Thus $lim_{x \to c} f (x) = f (c)$ so the limit exists and is as we expect.

☐

Similar with the Algebraic Limit Theorem for Functional Limits, we can do the same for derivatives:

Algebraic Differentiability Theorem

Let $f, g : I \to R$ where $I$ is an interval, $c \in I$ . Suppose $f, g$ are both differentiable at $c$ . Then $\forall α \in R$ then:

$α f$
$f + g$
$f g$
$\frac{f}{g}$ (assuming $g (c) \neq 0$ )
Are all differentiable at $c$ . Further we can say:
$(α f)^{'} (c) = α f^{'} (c)$
$(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$
$(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$
If $g (c) \neq 0$ then:
${(\frac{f}{g})}^{'} (c) = \frac{f^{'} (c) g (c) - f (c) g^{'} (c)}{g (c)^{2}}$

For a lot of these we'll be using the Algebraic Limit Theorem for Functional Limits

Proof

$\begin{aligned} (α f)^{'} (c) & = lim_{x \to c} \frac{(α f) (x) - (α f) (c)}{x - c} \\ = lim_{x \to c} \frac{α f (x) - α f (c)}{x - c} \\ = lim_{x \to c} (\frac{α f (x) - f (c)}{x - c}) \\ = α lim_{x \to c} \frac{f (x) - f (c)}{x - c} & ALT \\ = α f^{'} (c) \end{aligned}$
2.
$\begin{aligned} (f + g)^{'} (c) & = lim_{x \to c} \frac{(f + g) (x) - (f + g) (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) + g (x) - f (x) - g (x)}{x - c} \\ = lim_{x \to c} (\frac{f (x) - f (c)}{x - c} + \frac{g (x) - g (c)}{x - c}) \\ = lim_{x \to c} \frac{f (x) - f (c)}{x - c} + lim_{x \to c} \frac{g (x) - g (c)}{x - c} & ALT \\ = f^{'} (c) + g^{'} (c) \end{aligned}$
3. Notice here that for the $lim_{x \to c} g (x) = g (c)$ because Differentiability Implies Continuity:
$\begin{aligned} (f g)^{'} (c) & = lim_{x \to c} \frac{(f g) (x) - (f g) (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) g (x) - f (c) g (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) g (x) - f (c) g (x) + g (x) f (c) - g (c) f (c)}{x - c} \\ = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \cdot g (x) + \frac{g (x) - g (c)}{x - c} \cdot f (c) \\ = lim_{x \to c} \frac{f (x) - f (c)}{x - c} \cdot g (x) + lim_{x \to c} \frac{g (x) - g (c)}{x - c} \cdot f (c) & ALT \\ = lim_{x \to c} (\frac{f (x) - f (c)}{x - c}) lim_{x \to c} g (x) + lim_{x \to c} (\frac{g (x) - g (c)}{x - c}) lim_{x \to c} f (c) & ALT \\ = f^{'} (c) g (c) + g^{'} (c) f (c) \end{aligned}$
4. Assuming $g (c) \neq 0$ :
$\begin{aligned} {(\frac{f}{g})}^{'} (c) & = lim_{x \to c} (\frac{(\frac{f}{g}) (x) - (\frac{f}{g} (c))}{x - c}) \\ = lim_{x \to c} \frac{\frac{f (x)}{g (x)} - \frac{f (c)}{g (c)}}{x - c} \\ = lim_{x \to c} \frac{f (x) g (c) - f (c) g (x)}{g (x) g (c) (x - c)} \\ = lim_{x \to c} \frac{f (x) g (c) - f (c) g (c) + f (c) g (c) - f (c) g (x)}{g (x) g (c) (x - c)} \\ = lim_{x \to c} \frac{1}{g (x) g (c)} \cdot (\frac{f (x) - f (c)}{x - c} \cdot g (c) - f (x) \frac{g (x) - g (c)}{x - c}) \\ = lim_{x \to c} \frac{1}{g (x) g (c)} \cdot (f^{'} (c) g (c) - f (c) g^{'} (c)) & ALT \\ = \frac{f^{'} (c) g (c) - f (c) g^{'} (c)}{g (c)^{2}} \end{aligned}$

☐

Examples

Let's look at some quick examples of derivatives we could find:

(constant functions): $f (x) = α$ where $α \in R$ gives that:

lim_{x \to c} \frac{f (x) - f (c)}{x - c} = lim_{x \to c} \frac{α - α}{x - c} = 0

For $f : R \to R$ and $f (x) = x$ then:

lim_{x \to c} \frac{f (x) - f (c)}{x - c} = lim_{x \to c} \frac{x - c}{x - c} = 1

If $g (x) = x^{2} = x \cdot x$ then using Algebraic Differentiability Theorem:

g^{'} (x) = 1 \cdot x + x \cdot 1 = 2 x

Using induction, we can show that $h (x) = x^{n}$ where $n \in N$ then:

h^{'} (x) = n x^{n - 1}

Using (4) and the ADT, then any polynomial is differentiable on $R$ .
Further, any rational function is differentiable everywhere it is defined using (5) and the ADT (the quotient rule).