# This Quantum World/Appendix/Calculus

#### Differential calculus: a very brief introduction

Another method by which we can obtain a well-defined, finite number from infinitesimal quantities is to divide one such quantity by another.

We shall assume throughout that we are dealing with well-behaved functions, which means that you can plot the graph of such a function without lifting up your pencil, and you can do the same with each of the function's derivatives. So what is a function, and what is the derivative of a function?

A function ${\displaystyle f(x)}$ is a machine with an input and an output. Insert a number ${\displaystyle x}$ and out pops the number ${\displaystyle f(x).}$ Rather confusingly, we sometimes think of ${\displaystyle f(x)}$ not as a machine that churns out numbers but as the number churned out when ${\displaystyle x}$ is inserted.

The (first) derivative ${\displaystyle f'(x)}$ of ${\displaystyle f(x)}$ is a function that tells us how much ${\displaystyle f(x)}$ increases as ${\displaystyle x}$ increases (starting from a given value of ${\displaystyle x,}$ say ${\displaystyle x_{0}}$) in the limit in which both the increase ${\displaystyle \Delta x}$ in ${\displaystyle x}$ and the corresponding increase ${\displaystyle \Delta f=f(x+\Delta x)-f(x)}$ in ${\displaystyle f(x)}$ (which of course may be negative) tend toward 0:

${\displaystyle f'(x_{0})=\lim _{\Delta x\rightarrow 0}{\Delta f \over \Delta x}={df \over dx}(x_{0}).}$

The above diagrams illustrate this limit. The ratio ${\displaystyle \Delta f/\Delta x}$ is the slope of the straight line through the black circles (that is, the ${\displaystyle \tan }$ of the angle between the positive ${\displaystyle x}$ axis and the straight line, measured counterclockwise from the positive ${\displaystyle x}$ axis). As ${\displaystyle \Delta x}$ decreases, the black circle at ${\displaystyle x+\Delta x}$ slides along the graph of ${\displaystyle f(x)}$ towards the black circle at ${\displaystyle x,}$ and the slope of the straight line through the circles increases. In the limit ${\displaystyle \Delta x\rightarrow 0,}$ the straight line becomes a tangent on the graph of ${\displaystyle f(x),}$ touching it at ${\displaystyle x.}$ The slope of the tangent on ${\displaystyle f(x)}$ at ${\displaystyle x_{0}}$ is what we mean by the slope of ${\displaystyle f(x)}$ at ${\displaystyle x_{0}.}$

So the first derivative ${\displaystyle f'(x)}$ of ${\displaystyle f(x)}$ is the function that equals the slope of ${\displaystyle f(x)}$ for every ${\displaystyle x.}$ To differentiate a function ${\displaystyle f}$ is to obtain its first derivative ${\displaystyle f'.}$ By differentiating ${\displaystyle f',}$ we obtain the second derivative ${\displaystyle f''={\frac {d^{2}f}{dx^{2}}}}$ of ${\displaystyle f,}$ by differentiating ${\displaystyle f''}$ we obtain the third derivative ${\displaystyle f'''={\frac {d^{3}f}{dx^{3}}},}$ and so on.

It is readily shown that if ${\displaystyle a}$ is a number and ${\displaystyle f}$ and ${\displaystyle g}$ are functions of ${\displaystyle x,}$ then

${\displaystyle {d(af) \over dx}=a{df \over dx}}$  and  ${\displaystyle {d(f+g) \over dx}={df \over dx}+{dg \over dx}.}$

A slightly more difficult problem is to differentiate the product ${\displaystyle e=fg}$ of two functions of ${\displaystyle x.}$ Think of ${\displaystyle f}$ and ${\displaystyle g}$ as the vertical and horizontal sides of a rectangle of area ${\displaystyle e.}$ As ${\displaystyle x}$ increases by ${\displaystyle \Delta x,}$ the product ${\displaystyle fg}$ increases by the sum of the areas of the three white rectangles in this diagram:

In other "words",

${\displaystyle \Delta e=f(\Delta g)+(\Delta f)g+(\Delta f)(\Delta g)}$

and thus

${\displaystyle {\frac {\Delta e}{\Delta x}}=f\,{\frac {\Delta g}{\Delta x}}+{\frac {\Delta f}{\Delta x}}\,g+{\frac {\Delta f\,\Delta g}{\Delta x}}.}$

If we now take the limit in which ${\displaystyle \Delta x}$ and, hence, ${\displaystyle \Delta f}$ and ${\displaystyle \Delta g}$ tend toward 0, the first two terms on the right-hand side tend toward ${\displaystyle fg'+f'g.}$ What about the third term? Because it is the product of an expression (either ${\displaystyle \Delta f}$ or ${\displaystyle \Delta g}$) that tends toward 0 and an expression (either ${\displaystyle \Delta g/\Delta x}$ or ${\displaystyle \Delta f/\Delta x}$) that tends toward a finite number, it tends toward 0. The bottom line:

${\displaystyle e'=(fg)'=fg'+f'g.}$

This is readily generalized to products of ${\displaystyle n}$ functions. Here is a special case:

${\displaystyle (f^{n})'=f^{n-1}\,f'+f^{n-2}\,f'\,f+f^{n-3}\,f'\,f^{2}+\cdots +f'\,f^{n-1}=n\,f^{n-1}f'.}$

Observe that there are ${\displaystyle n}$ equal terms between the two equal signs. If the function ${\displaystyle f}$ returns whatever you insert, this boils down to

${\displaystyle (x^{n})'=n\,x^{n-1}.}$

Now suppose that ${\displaystyle g}$ is a function of ${\displaystyle f}$ and ${\displaystyle f}$ is a function of ${\displaystyle x.}$ An increase in ${\displaystyle x}$ by ${\displaystyle \Delta x}$ causes an increase in ${\displaystyle f}$ by ${\displaystyle \Delta f\approx {\frac {df}{dx}}\Delta x,}$ and this in turn causes an increase in ${\displaystyle g}$ by ${\displaystyle \Delta g\approx {\frac {dg}{df}}\Delta f.}$ Thus ${\displaystyle {\frac {\Delta g}{\Delta x}}\approx {\frac {dg}{df}}{\frac {df}{dx}}.}$ In the limit ${\displaystyle \Delta x\rightarrow 0}$ the ${\displaystyle \approx }$ becomes a ${\displaystyle =}$ :

${\displaystyle {dg \over dx}={dg \over df}{df \over dx}.}$

We obtained ${\displaystyle (x^{n})'=n\,x^{n-1}}$ for integers ${\displaystyle n\geq 2.}$ Obviously it also holds for ${\displaystyle n=0}$ and ${\displaystyle n=1.}$

1. Show that it also holds for negative integers ${\displaystyle n.}$ Hint: Use the product rule to calculate ${\displaystyle (x^{n}x^{-n})'.}$
2. Show that ${\displaystyle ({\sqrt {x}})'=1/(2{\sqrt {x}}).}$ Hint: Use the product rule to calculate ${\displaystyle ({\sqrt {x}}{\sqrt {x}})'.}$
3. Show that ${\displaystyle (x^{n})'=n\,x^{n-1}}$ also holds for ${\displaystyle n=1/m}$ where ${\displaystyle m}$ is a natural number.
4. Show that this equation also holds if ${\displaystyle n}$ is a rational number. Use ${\displaystyle {dg \over dx}={dg \over df}{df \over dx}.}$

Since every real number is the limit of a sequence of rational numbers, we may now confidently proceed on the assumption that ${\displaystyle (x^{n})'=n\,x^{n-1}}$ holds for all real numbers ${\displaystyle n.}$