This Quantum World/Appendix/Calculus

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Differential calculus: a very brief introduction[edit]

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.


Derivative1.png


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:


Product.png


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.