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

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

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

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

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

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

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

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

In other "words",

$\Delta e=f(\Delta g)+(\Delta f)g+(\Delta f)(\Delta g)$

and thus

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

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

$e'=(fg)'=fg'+f'g.$

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

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

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

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

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

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

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

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

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