# 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 ${\displaystyle n.}$