# 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 $A\subset ℝ$.

We call a function  $f:A\to ℝ$  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:  f-prime at a)

 ${f\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)≔\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}\left(x\right)=\underset{x\to a}{\mathrm{lim}}\frac{f\left(x\right)-f\left(a\right)}{x-a}$ [7.3.1]

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

Consider:

• Being the limit of secant gradients we regard  ${f\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)$ as the gradient of the tangent at  f  in (a, f(a)).

• As a is a accumulation point of A - and thus also of $A\\left\{a\right\}$ -   ${m}_{\phantom{\rule{0.0em}{0ex}}a}$ has at most one continuous extension at a. Therefor the limit at a, i.e. the number  ${f\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)$ is uniquely determined.

• The symbol  ${f}^{\prime }\left(a\right)$ is due to Cauchy. Although quite common it is not seldom replaced by Leibniz's notation $\frac{d\phantom{\rule{1px}{0ex}}f}{dx}\left(a\right)$ and $\frac{d}{dx}\phantom{\rule{0.2em}{0ex}}f\left(a\right)$ respectively. The idea behind is, that when "performing the limit process" the quotient of the differences $\mathrm{\Delta }\phantom{\rule{0.2em}{0ex}}f=f\left(x\right)-f\left(a\right)$ and $\mathrm{\Delta }\phantom{\rule{0.2em}{0ex}}x=x-a$ 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 $t↦s\left(t\right)$ type (cf. the note in [7.2]) the derivative number is denoted by a dot symbol

$\stackrel{\mathbf{˙}}{s}\left({t}_{0}\right)=\underset{t\to {t}_{0}}{\mathrm{lim}}\frac{\mathrm{\Delta }\phantom{\rule{0.1em}{0ex}}s}{\mathrm{\Delta }\phantom{\rule{0.1em}{0ex}}t}=\underset{t\to {t}_{0}}{\mathrm{lim}}\frac{s\left(t\right)-s\left({t}_{0}\right)}{t-{t}_{0}}$

and referred to as the instantaneous speed at the time ${t}_{0}$. It is usually replaced by  $v\left({t}_{0}\right)=\stackrel{\mathbf{˙}}{s}\left({t}_{0}\right)$.

It is this textual concept that the alternative term "instantaneous rate of change of the function  f at a" for the derivative  ${f}^{\prime }\left(a\right)$ comes from.

Example:
• Every linear function  $mX+b$  is differentiable at each $a\in ℝ$ with
 ►   $\left(mX+b{\right)}^{\prime }\left(a\right)=m$ [7.3.2]

Proof: According [7.2.2]  ${m}_{\phantom{\rule{0.0em}{0ex}}a}$ is continuously extendable at a by the constant function m on $ℝ$. Thus we have:

$\left(mX+b{\right)}^{\prime }\left(a\right)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}\left(x\right)=m\left(a\right)=m$.

• With $n\in ℕ$ the power function  ${X}^{n}$  is differentiable at every $a\in ℝ$ and
 ►   $\left({X}^{n}{\right)}^{\prime }\left(a\right)=n{a}^{n-1}$ [7.3.3]

Proof: From [7.2.3] we see that  ${m}_{\phantom{\rule{0.0em}{0ex}}a}$ is continuously extendable at a by $\sum _{i=0}^{n-1}{a}^{i}{X}^{n-i-1}$. From this we calculate the derivative number:

$\left({X}^{n}{\right)}^{\prime }\left(a\right)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}\left(x\right)=\sum _{i=0}^{n-1}{a}^{i}{X}^{n-i-1}\left(a\right)=\sum _{i=0}^{n-1}{a}^{i}{a}^{n-i-1}=n{a}^{n-1}$.

• The reciprocal function  $\frac{1}{X}$ is differentiable at every $a\ne 0$ with
 ►   $\left(\frac{1}{X}{\right)}^{\prime }\left(a\right)=-\frac{1}{{a}^{2}}$ [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

$\left(\frac{1}{X}{\right)}^{\prime }\left(a\right)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}\left(x\right)=-\frac{1}{aX}\left(a\right)=-\frac{1}{{a}^{2}}$ .

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

${\sqrt{X}\phantom{\rule{0.1em}{0ex}}}^{\prime }\left(a\right)=\underset{x\to a}{\mathrm{lim}}{m}_{\phantom{\rule{0.0em}{0ex}}a}\left(x\right)=\frac{1}{\sqrt{X}+\sqrt{a}}\left(a\right)=\frac{1}{2\sqrt{a}}$

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 $\left(\frac{1}{{n}^{2}}\right)$, a sequence in ${ℝ}^{>0}$ converging to 0, with the image sequence $\left({m}_{0}\left(\frac{1}{{n}^{2}}\right)\right)=\left(n\right)$ 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 $\left(\frac{{\left(-1\right)}^{n}}{n}\right)$ in ${ℝ}^{\ne 0}$. The image sequence $\left({\left(-1\right)}^{n}\right)$ is divergent.

Consider:

• According [7.3.2] especially all the constant functions c and the identity function X as well are differentiable everywhere with the following derivative numbers:
 $\begin{array}{c}{c\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)=0\\ {X\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)=1\text{.}\end{array}$ [7.3.6]

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   $f:A\to ℝ$ be differentiable at $a\in A$. We call the function

 ${t}_{a}≔f\left(a\right)+\left(X-a\right){f\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)$ [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 ${t}_{a}\left(a\right)=f\left(a\right)$, and that ${f\phantom{\rule{0.05em}{0ex}}}^{\prime }\left(a\right)$ is the gradient factor of  ${t}_{a}$.

• Independently from A tangent functions always have the whole of $ℝ$ 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 non-vertical normals:

Definition:  If   $f:A\to ℝ$ is differentiable at $a\in A$ such that ${f}^{\prime }\left(a\right)\ne 0$ the function

 ${n}_{a}≔f\left(a\right)-\left(\mathrm{X}-a\right)\frac{1}{{f}^{\prime }\left(a\right)}$ [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 ${n}_{a}\left(a\right)=f\left(a\right)$  ${n}_{a}$  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 non-zero 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 $ℝ$ 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 ${\mathrm{X}}^{3}\left(1\right)=1$ and $\left({\mathrm{X}}^{3}{\right)}^{\prime }\left(1\right)=3\cdot {1}^{2}=3$ we have $\begin{array}{l}{t}_{1}=1+\left(\mathrm{X}-1\right)\cdot 3=3\mathrm{X}-2\\ {n}_{1}=1-\left(\mathrm{X}-1\right)\cdot \frac{1}{3}=-\frac{1}{3}\mathrm{X}+\frac{4}{3}\end{array}$

 Exercise:  Calculate the tangent and normal function of $\sqrt{X}$ with respect to 4: $\phantom{|}{t}_{4}={?}\sqrt{4}+\left(X-4\right)\frac{1}{2\sqrt{4}}=\frac{1}{4}X+1$ $\phantom{|}{n}_{4}={?}\sqrt{4}-\left(X-4\right)\cdot 2\sqrt{4}=-4X+18$ $\frac{1}{X}$ with respect to −1: $\phantom{|}{t}_{-1}={?}\frac{1}{-1}+\left(X+1\right)\left(-\frac{1}{{\left(-1\right)}^{2}}\right)=-X-2$ $\phantom{|}{n}_{-1}={?}\frac{1}{-1}-\left(X+1\right)\left(-{\left(-1\right)}^{2}\right)=X$ $mX+b$ with respect to a:

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, (af(a)) is the most obvious one, and to calculate a directional vector. Having in mind that for an horizontal increment of 1 the derivative ${f}^{\prime }\left(a\right)$ is the appropriate vertical increment we find our vector representation as follows:

 ${t}_{a}=\mathrm{\left(}\begin{array}{c}a\\ f\left(a\right)\end{array}\mathrm{\right)}+<\mathrm{\left(}\begin{array}{c}1\\ {f}^{\prime }\left(a\right)\end{array}\mathrm{\right)}>$ [7.3.9]

As both vectors  $\mathrm{\left(}\begin{array}{c}{f}^{\prime }\left(a\right)\\ -1\end{array}\mathrm{\right)}$  and  $\mathrm{\left(}\begin{array}{c}1\\ {f}^{\prime }\left(a\right)\end{array}\mathrm{\right)}$  are perpendicular to each other the normal could be represented as

 ${n}_{a}=\mathrm{\left(}\begin{array}{c}a\\ f\left(a\right)\end{array}\mathrm{\right)}+<\mathrm{\left(}\begin{array}{c}{f}^{\prime }\left(a\right)\\ -1\end{array}\mathrm{\right)}>$ [7.3.10]

[7.3.10] has the additional advantage that vertical normals are included as well. Consider however that in the non-vertical case $\mathrm{\left(}\begin{array}{c}{f}^{\prime }\left(a\right)\\ -1\end{array}\mathrm{\right)}$  and  $\mathrm{\left(}\begin{array}{c}1\\ -1}{{f}^{\prime }\left(a\right)}\end{array}\mathrm{\right)}$ 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 $\mathrm{\left(}\begin{array}{c}1\\ 1\end{array}\right)+<\left(\begin{array}{c}3\\ -1\end{array}\mathrm{\right)}>$.

 7.2. 7.4.