On-line Math 21

2.2  Differentiation Rules

Theorem 7 [The Chain Rule]. ( f(g(x))) ¢ = f¢(g(x))·g¢(x).

Proof:

Again, we appeal to the definition of the derivative:

( f(g(x))) ¢ : = lim h® 0  f(g(x+h))-f(g(x)) h = lim h® 0  (f(g(x+h))-f(g(x)))(g(x+h)-g(x)) h(g(x+h)-g(x)) .

Now, we do some clever substitutions; some may say sleight-of-hand substitutions. The idea here is to get you to look at the same term two different ways. Substitute u = g(x) , and u+Du = g(x+h) (so that Du = g(x+h)-g(x) ), and:

( f(g(x))) ¢ = lim h® 0  (f(g(x+h))-f(g(x)))(g(x+h)-g(x)) h(g(x+h)-g(x)) = lim h® 0  (f(u+Du))-f(u))(g(x+h)-g(x)) h(Du) = æ ç è lim Du® 0  f(u+Du))-f(u) Du ö ÷ ø æ ç è lim h® 0  g(x+h)-g(x)) h ö ÷ ø = f¢(u)·g¢(x) = f¢(g(x))·g¢(x).

Note how you are looking at Du in the first difference quotient as the change in the variable, but in the second difference quotient it changes roles to the difference of the values of g . Now, also, I changed the first limit to as Du® 0 . This is justifiable, since as h® 0 , Du® 0 .

There is one thing wrong with this proof, though, and that is that it might be that g(x+h) = g(x) lots of times as h goes down to 0, so the limits as written won't make much sense. That is, while certainly Du® 0 when h® 0 , it might be that Du goes to 0 well before h does, so you can't quite say that h® 0 when Du® 0 . However, if that happens, that is, if Du goes to zero before h does, or hits 0 many times as h® 0 , then g¢(x) = 0 , as will be ( f(g(x))) ¢, so the formula works even in this case. There is another proof which works around this difficulty, but this is the proof that makes the most sense, despite this theoretical question mark.

Copyright (c) 2000 by David L. Johnson.