A simple proof of the chain rule

In the textbook for the japanese senior high school students, the chain rule is proved from the formula Δz/Δx = (Δz/Δy)(Δy/Δx). If Δx → 0, then we have dz/dx = (dz/dy)(dy/dx). But this proof is not correct, because Δx ≠ 0 does not imply Δy ≠ 0.

It seems that I have a simple proof of the chain rule. The proof is as follows:

Assume that y = f(x), z = g(y), and that f and g are differentiable at x0, f(x0) respectively. Put Δy = f(x0+Δx) - f(x0) and Δz = (gf)(x0+Δx) - gf(x0).

If f′(x0) ≠ 0, then Δy ≠ 0 for sufficiently small Δx (≠ 0). Therefore the argument above is correct.

If f′(x0) = 0, then we have to show that (gf)′(x0) = 0.

Therefore Δz/Δx → 0.

(More precisely:

There exists a positive number δ1 and a positive number M such that 0 < |Δy| < δ1 implies |Δz/Δy| < M. Therefore |Δy| < δ1 implies Δz = 0 or |Δz/Δy| < M.

There exists a positive number δ2 such that |Δx| < δ2 implies |Δy| < δ1. Therefore |Δx| < δ2 implies Δz = 0 or |Δz/Δy| < M.

For any positive number ε, there exists a positive number δ3 such that 0 < |Δx| < δ3 implies |Δy/Δx| < ε/M. Put δ = min(δ2, δ3). Then 0 < |Δx| < δ implies Δz = 0 or |(Δz/Δy)(Δy/Δx)| < M * ε / M = ε.)


Iwase Zjuñici