# 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:

$f'(x_0)=\lim_{\Delta x\rightarrow0}{\Delta f\over\Delta x}={df\over dx}(x_0).$

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 $f''=\frac{d^2f}{dx^2}$ of $f,$ by differentiating $f''$ we obtain the third derivative $f'''=\frac{d^3f}{dx^3},$ 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 $\Delta x\rightarrow0$ the $\approx$ becomes a $=$ :

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

We obtained $(x^n)'=n\,x^{n-1}$ for integers $n\geq2.$ 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 $(x^nx^{-n})'.$
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 $n.$