Expert: Daniel Mazur - 5/14/2009

Question
Hi daniel,
I know i should be asking this in the mathematics section, but i don't want formal mathematical definitions that i already know, and want to understand it the way a physicist would like to.
Ok, first of all, i'm clear with the concept of limits and all and am clear with basic calculus.

Now i recently read the e-book, calculus for the practical man by j.e. thompson. this is the book that feynman claimed to have studied calculus from for the first time.
now the author presents it without the concept of limits.
so dy and dx in this book are used for a change, not necessarily an infinitesimelly small change.

Now the problem is that it states, that "if y is a func. of x, then the change in y i.e. dy will dpend on the change in x i.e. dx and 'the value of x itself' ".

It can be understood logically that since y is a func. of x, so the change in value of y will depend on the change in the value of x. what i fail to understand is that how can it be logically deduced that the change in the value of y should also depend on x. value of y at a particular point certainly depends on the value of x at that point but the change in y doesn't depend on the value of x?
and even if i accept that it does, then on which value of x does it depend? if the change takes place over a certain interval, there are infinite x's in this interval, so on which x does the change in y also depend?(remember dy or dx in not infinitesimally small).

also, suppose we abandon the above general idea of dy and dx and go back to the infinitesimal small changes dy and dx. And, now suppose we have a function, say y=x^2, a parabola, then it can be seen from the graph also, that the same change in x produces a diff. change in y at diff. points.so in this case a change in y will also depend on x. also, now let's take the change in x so small i.e.dx, that the propotion by which y changes, is fixed for this small interval dx and does not vary much with points contained inside the interval dx itself(dx is infinitesimally small). then we know that the propotion by which y changes will be 2x (differentiating y=x^2) in this interval where x is either the starting or ending point of the small interval dx. but how can this result be derived in a non-mathematical way? how do we know that the propotion by wich y changes is 2x. it is understood that the propotion by which it changes will be different for different x, from the graph, but how do we know that it's exactly 2x. if you could explain it in as non-mathematical a way as possible, it would be great.

Thanks,
Shikhin

Answer
Hi Shikhin,

I think that it is a matter of a point of view and the extent to which we can non-mathematically reason out mathematical formulae. Firstly, why from dependence y(x) also follows dy=f(x)dx? I don't think that whether you take the infinitesimal reason or not, the answer should not be different, because our experience says that infinitesimal is a very straightforward limit of finite element calculus.

Let us say that in truth dy (taken at constant dx!) does NOT depend on x. This would be the case only if dy/dx was a constant. This can be disproved by counterexample: y=x^2 does not offer a constant dy/dx. So, our saying that dy depends on x is only another way of saying that "dy is not constant". This last statement is our everyday observation - and that is also my non-mathematical reasoning.

Secondly, you wish to know, how we choose the x, on which dy depends. Well, function y(x) is nothing but an image of all points of x (from the range, where y(x) is defined) taken one after another. When we calculate a derivative, it is generally just another function applied on the result (!) values of the function y(x), so we could easily mark the derivative something like G(y):=G(y(x)). In this manner any chosen x0 from the range of x's gets produces one particular y0:=y(x0) and that in turn produces one particular G0:=G(y0)=G(y(x0))=G(x0). There can be no confusion as to which x0 to pick, you take for every function result the value of argument that had produced it! Suppose y=x^2. For x0=1 the y0=x0^2=1 and the G0=2*x0=2. For another x0 the same relations will hold.

Lastly, how do we find that the first derivative of y=x^2 is y'=G=2*x? This is done by starting with definition of finite differences, their processing and finally making a limiting step from finite to infinitesimal. As I had my maths course including limits and as I am no mathematician, it would be beyond my expertise to say, how to derive y'=2*x without the crucial limiting step. The following I am taking from http://en.wikipedia.org/wiki/Derivative:

Take for example x0=3 in function y=x^2. Definition of derivative (dx and dy mean finite differencesfor now) at a point x0 can be written as
dy/dx(x0) = [y(x0+dx)-y(x)]/dx = [(3+dx)^2-3^2]/dx = [3^2+2*3*dx+dx^2]/dx = [6*dx+dx^2]/dx = 6+dx
The last expression, in the limit dx->0, results in dy/dx(3) = 6 = 2*x0. As y'(x0)=2*x0 for any other x0 as well, we can write simply y'=2x for y=x^2. That's how this has been calculated.

As a next step you can make a more general calculation for all polynomials and that will give you the reason for the general rule that y'=n*x^(n-1) for y=x^n. The pre-factor comes from the second binomial term of (x+dx)^n, there is no mystery.

That should clarify things a bit, I am glad if I helped.
Daniel