Proof of the Chain Rule

Date: 09/06/2004 at 10:13:43
From: Philip
Subject: Chain rule proof

I am using the chain rule (dy/dx = dy/du * du/dx) in my math class, 
and I would like to see it proved, which we don't do in class.

My teacher told me the formal proof is an epsilon-delta proof, and in 
my spare time I have studied that kind of proof a little (using your 
splendid archives) so I can understand this proof.

Date: 09/11/2004 at 10:22:58
From: Doctor Fenton
Subject: Re: Chain rule proof

Hi Philip,

Thanks for writing to Dr. Math.  While epsilons and deltas are
necessary to prove the technical details, you may already have
accepted the necessary limit theorems for a proof: the limit of a
product is the product of the limits; and if a function f is
continuous at y = g(a), while the function g is continuous at a, then
f(g(x)) is continuous at x = a (assuming that the composite function
f(g(x)) is defined in some open interval containing x = a).

The Chain Rule applies to composite functions f(g(x)), when g is
differentiable (and therefore continuous) at x = a, and f is 
differentiable at y = g(a).  To investigate differentiability, the 
first thing to do is to examine the difference quotient:

   f(g(x)) - f(g(a))
   -----------------
         x - a         ,

knowing that

        f(y) - f(g(a))
   lim  -------------- = f'(g(a))
  y->g(a)  y - g(a)

and
        g(x) - g(a)
   lim  -------------- = g'(a)    .
  x->a     x - a

The "natural" approach is to write

   f(g(x)) - f(g(a))   f(g(x)) - f(g(a))   g(x) - g(a)
   ----------------- = ----------------- * -----------
         x - a            g(x) - g(a)         x - a

and take the limit of both sides.  However, there are functions (such 
as g(x) = x^2 sin(1/x) for a = 0) for which g(x) - g(a) is 0 
infinitely often as x->a.  This behavior cannot happen if g'(a) is 
non-zero, so this proof would work in that case.  However, if you want 
to cover the case g'(a) = 0 as well, you have to be more careful.

Let 
          f(y) - f(g(a))
   F(y) = --------------    for y different from g(a) ,
             y - g(a)

and define  F(g(a)) = f'(g(a)) .  Then F(y) is continuous at y = g(a), 
so F(g(x)) is continuous at x = a.  Then we can write

   f(g(x)) - f(g(a))             g(x) - g(a)
   ----------------- = F(g(x)) * -----------
         x - a                      x - a

and this equation is true even if g(x) = g(a), since both sides are 0 
(and the right side is not undefined, as before).

Now, take the limit as x->a, and you obtain the Chain Rule.

If you have any questions, please write back and I will try to explain 
further.

- Doctor Fenton, The Math Forum
  http://mathforum.org/dr.math/