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 $$n.$$