7.3. Differentiable Functions
Again we take the result of 7.1. as our guideline. We have already transferred the spectrum of secant gradients into a mathematical notion,
so the remaining task will be to deal with the tangent gradient. We take [7.1.2] as a pattern.
Definition: Let $a\in A$
be an accumulation point of $A\subset \mathbb{R}$.
We call a function $f:A\to \mathbb{R}$ differentiable at a, if the difference quotient function $m}_{\phantom{\rule{0.0em}{0ex}}a$
is continuously extendable at a. In this case the real number (read: fprime at a)
$f\phantom{\rule{0.05em}{0ex}}}^{\prime}(a)\u2254\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}(x)=\underset{x\to a}{\mathrm{lim}}\frac{f(x)f(a)}{xa$

[7.3.1] 
is called derivative (more precise: derivative number) of f at a.

Consider:
Being the limit of secant gradients we regard ${f\phantom{\rule{0.05em}{0ex}}}^{\prime}(a)$
as
the gradient of the tangent at f in (a, f(a)).
As a is a accumulation point of A  and thus also of $A\backslash \left\{a\right\}$
 $m}_{\phantom{\rule{0.0em}{0ex}}a$ has at most one continuous extension at a. Therefor the limit at a, i.e. the
number ${f\phantom{\rule{0.05em}{0ex}}}^{\prime}(a)$ is uniquely determined.
The symbol ${f}^{\prime}(a)$ is due to Cauchy. Although quite common it is not seldom replaced by Leibniz's notation $\frac{d\phantom{\rule{1px}{0ex}}f}{dx}(a)$ and $\frac{d}{dx}\phantom{\rule{0.2em}{0ex}}f(a)$ respectively. The idea behind is, that when "performing the limit process" the quotient of the differences $\mathrm{\Delta}\phantom{\rule{0.2em}{0ex}}f=f(x)f(a)$ and $\mathrm{\Delta}\phantom{\rule{0.2em}{0ex}}x=xa$ will switch over into the
mystic
i 
Note that there was no precise limit idea for Leibniz (1646  1716) and it was not until Cauchy (1789  1857) who developed a modern, satisfactory concept.

quotient of differentials df and dx. We read $\frac{d\phantom{\rule{1px}{0ex}}f}{dx}$ as "dee eff dee ecks" to indicate, at least when reading, that there is no real quotient.

Again physics has its own notation at this point: With time related functions, e.g. with those of the $t\mapsto s(t)$ type (cf. the note in [7.2]) the derivative number is denoted by a dot symbol
$\dot{s}({t}_{0})=\underset{t\to {t}_{0}}{\mathrm{lim}}\frac{\mathrm{\Delta}\phantom{\rule{0.1em}{0ex}}s}{\mathrm{\Delta}\phantom{\rule{0.1em}{0ex}}t}=\underset{t\to {t}_{0}}{\mathrm{lim}}\frac{s(t)s({t}_{0})}{t{t}_{0}}$
and referred to as the instantaneous speed at the time $t}_{0$. It is usually replaced by $v({t}_{0})=\dot{s}({t}_{0})$.
It is this textual concept that the alternative term "instantaneous rate of change of the function f at a" for the derivative ${f}^{\prime}(a)$ comes from.
Example:

Every linear function $mX+b$ is differentiable at each $a\in \mathbb{R}$
with
Proof: According [7.2.2] $m}_{\phantom{\rule{0.0em}{0ex}}a$ is continuously extendable at a by the constant function m on $\mathbb{R}$.
Thus we have:
$(mX+b{)}^{\prime}(a)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}(x)=m(a)=m$.

With $n\in \mathbb{N}$ the power function ${X}^{n}$ is differentiable at every $a\in \mathbb{R}$ and
►
$({X}^{n}{)}^{\prime}(a)=n{a}^{n1}$  [7.3.3]

Proof: From [7.2.3] we see that $m}_{\phantom{\rule{0.0em}{0ex}}a$ is continuously extendable at a by $\sum _{i=0}^{n1}{a}^{i}{X}^{ni1}$. From this we calculate the derivative number:
$({X}^{n}{)}^{\prime}(a)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}(x)=\sum _{i=0}^{n1}{a}^{i}{X}^{ni1}(a)=\sum _{i=0}^{n1}{a}^{i}{a}^{ni1}=n{a}^{n1}$.

The reciprocal function $\frac{1}{X}$ is differentiable at every $a\ne 0$ with
►
$(\frac{1}{X}{)}^{\prime}(a)=\frac{1}{{a}^{2}}$  [7.3.4] 
Proof: According [7.2.4] $\frac{1}{aX}$
is the continuous extension of $m}_{\phantom{\rule{0.0em}{0ex}}a$ at a. So we get the following derivative number
$(\frac{1}{X}{)}^{\prime}(a)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}(x)=\frac{1}{aX}(a)=\frac{1}{{a}^{2}}$ .

The root function $\sqrt{X}$ is differentiable at every $a>0$
and
►
$\sqrt{X}\phantom{\rule{0.1em}{0ex}}}^{\prime}(a)=\frac{1}{2\sqrt{a}$

[7.3.5] 
Proof: [7.2.5] shows that $m}_{\phantom{\rule{0.0em}{0ex}}a$ is continuously extended at a by $\frac{1}{\sqrt{X}+\sqrt{a}}$,
yielding
$\sqrt{X}\phantom{\rule{0.1em}{0ex}}}^{\prime}(a)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}(x)=\frac{1}{\sqrt{X}+\sqrt{a}}(a)=\frac{1}{2\sqrt{a}$
as derivative number.

The root function $\sqrt{X}$ is not differentiable at 0. The difference quotient function $m}_{0}=\frac{1}{\sqrt{X}$ is not continuously extendable at 0. We see this from $(\frac{1}{{n}^{2}})$,
a sequence in ${\mathbb{R}}^{>0}$ converging to 0, with the image sequence $({m}_{0}(\frac{1}{{n}^{2}}))=(n)$
being divergent.

The absolute value function $X$ is not differentiable at 0: $\frac{X}{X}$
(cf. [7.2.6]) can't be continuously extended.
Take e.g. the zero sequence $(\frac{{(1)}^{n}}{n})$
in ${\mathbb{R}}^{\ne 0}$. The image sequence $({(1)}^{n})$
is divergent.

Consider:
We started our issue looking for tangents. Having set up the derivative numbers the main problem, getting the correct gradient, is now solved. So we are prepared to build tangents.
Definition:
Let $f:A\to \mathbb{R}$ be differentiable at $a\in A$.
We call the function
${t}_{a}\u2254f(a)+(Xa){f\phantom{\rule{0.05em}{0ex}}}^{\prime}(a)$

[7.3.7] 
the tangent function of f with respect to a.

Consider:
Occasionally we write $t}_{f,a$ instead of $t}_{\phantom{\rule{0.0em}{0ex}}a$ to point to the relation to f.
$t}_{a$ is a linear function, touching the graph of f in (a, f(a)). Note that ${t}_{a}(a)=f(a)$, and that ${f\phantom{\rule{0.05em}{0ex}}}^{\prime}(a)$
is the gradient factor of $t}_{a$.
Independently from A tangent functions always have the whole of $\mathbb{R}$ as their domain.
Working with tangents often comprises the search for the so called normal of f, i.e. the perpendicular to the tangent running through (a, f(a)). There is a linear function to represent nonvertical normals:
Definition: If $f:A\to \mathbb{R}$ is differentiable at $a\in A$ such that ${f}^{\prime}(a)\ne 0$ the function
$n}_{a}\u2254f(a)(\mathrm{X}a)\frac{1}{{f}^{\prime}(a)$

[7.3.8] 
is called the normal function of f with respect to a.

Consider:
If necessary we use the more detailed notation $n}_{f,a$ for the normal function.
As ${n}_{a}(a)=f(a)$ $n}_{a$ and f actually intersect at a. Its slope is the reciprocal of the slope of $t}_{a$ with a reversed sign. $n}_{a$ is thus
perpendicular
i 
Rotating a straight line g with nonzero slope $m=\frac{h}{l}\ne 0$ by 90° results in a perpendicular line $g}^{\perp$ with a slope of
$\frac{l}{h}=\frac{1}{m}$

to $t}_{a$.
Independently from A normal functions always have the whole of $\mathbb{R}$ as their domain.
As an example we calculate the tangent and normal function for the cubic function $\mathrm{X}}^{3$ with respect to 1. With ${\mathrm{X}}^{3}(1)=1$ and $({\mathrm{X}}^{3}{)}^{\prime}(1)=3\cdot {1}^{2}=3$ we have
$\begin{array}{l}{t}_{1}=1+(\mathrm{X}1)\cdot 3=3\mathrm{X}2\\ {n}_{1}=1(\mathrm{X}1)\cdot \frac{1}{3}=\frac{1}{3}\mathrm{X}+\frac{4}{3}\end{array}$


Exercise: Calculate the tangent and normal function of
$\sqrt{X}$ with respect to 4:
$\phantom{}{t}_{4}={?}\sqrt{4}+(X4)\frac{1}{2\sqrt{4}}=\frac{1}{4}X+1$
$\phantom{}{n}_{4}={?}\sqrt{4}(X4)\cdot 2\sqrt{4}=4X+18$

$\frac{1}{X}$ with respect to −1:
$\phantom{}{t}_{1}={?}\frac{1}{1}+(X+1)(\frac{1}{{(1)}^{2}})=X2$
$\phantom{}{n}_{1}={?}\frac{1}{1}(X+1)({(1)}^{2})=X$

$mX+b$ with respect to a:
$\phantom{}{t}_{a}={?}ma+b+(Xa)m=mX+b\text{[sic!]}$

It is an advantage to switch to the vector notation and represent a tangent as a line. We only need to know one point of that line, (a, f(a)) is the most obvious one, and to calculate a directional vector. Having in mind that for an horizontal increment of 1 the derivative ${f}^{\prime}(a)$ is the appropriate vertical increment we find our vector representation as follows:
${t}_{a}=\mathrm{\left(}\begin{array}{c}a\\ f(a)\end{array}\mathrm{\right)}+<\mathrm{\left(}\begin{array}{c}1\\ {f}^{\prime}(a)\end{array}\mathrm{\right)}>$

[7.3.9] 
As both vectors $\mathrm{\left(}\begin{array}{c}{f}^{\prime}(a)\\ 1\end{array}\mathrm{\right)}$ and $\mathrm{\left(}\begin{array}{c}1\\ {f}^{\prime}(a)\end{array}\mathrm{\right)}$ are perpendicular to each other the normal could be represented as
${n}_{a}=\mathrm{\left(}\begin{array}{c}a\\ f(a)\end{array}\mathrm{\right)}+<\mathrm{\left(}\begin{array}{c}{f}^{\prime}(a)\\ 1\end{array}\mathrm{\right)}>$

[7.3.10] 
[7.3.10] has the additional advantage that vertical normals are included as well. Consider however that in the nonvertical case $\mathrm{\left(}\begin{array}{c}{f}^{\prime}(a)\\ 1\end{array}\mathrm{\right)}$ and $\mathrm{\left(}\begin{array}{c}1\\ \raisebox{1ex}{$1$}\!\left/ \!\raisebox{1ex}{${f}^{\prime}(a)$}\right.\end{array}\mathrm{\right)}}$ represent the same direction.
For an example we return to $\mathrm{X}}^{3$. The tangent of $\mathrm{X}}^{3$ with respect to 1 is the line $\mathrm{\left(}\begin{array}{c}1\\ 1\end{array}\right)+<\left(\begin{array}{c}1\\ 3\end{array}\mathrm{\right)}>$ and its normal is $\mathrm{\left(}\begin{array}{c}1\\ 1\end{array}\right)+<\left(\begin{array}{c}3\\ 1\end{array}\mathrm{\right)}>$.
