So, here's a an attempted proof of the chain rule. I end up with the right formula at the end of the day, but I'm curious what the right way is to substitute a function with a best linear approximation to it inside limits like these when you're trying to find an explicit form for a derivative involving unknown functions.

$f$ and $g$ in the attempted proof below are called/evaluated twice, but with arguments that are very close to each other. Based on that, I replaced $f(x)$ and $g(x)$ with linear approximations centered around one of their arguments.

Is there a way to perform a substitution like this rigorously?

Using a slightly modified form of the definition of the derivative.

$$ x \cdot D(f \circ g) = \lim_{h \to 1}\frac{f(g(hx)) - f(g(x))}{h-1} $$

next, I use the following linear approximations to $f$ and $g$ at strategically chosen points. $$ f(z) \approx cz + d $$

$cz + d$ is the best linear approximation to $f$ at $(g(x), f(g(x))$

$$ g(z) \approx az + b $$

$az + b$ is the best linear approximation to $g$ at $(x, g(x))$

$$ \lim_{h \to 1} \frac{f(ahx + b) - f(ax+b)}{h-1} $$

$$ \lim_{h \to 1} \frac{achx + cb + d - cax - cb - d}{h - 1} $$

$$ \lim_{h \to 1} \frac{ca(h-1)x}{h-1} $$

$$ \lim_{h \to 1} cax $$

And thus:

$$ x \cdot D(f \circ g) = cax = (Df)(g(x)) \cdot (Dg)(x) \cdot x $$

$$ D(f \circ g) = (Df)(g(x)) \cdot (Dg)(x) $$

  • 1
    $\begingroup$ There’s a lot hidden in that $\approx$. Generally speaking, if you’re going to approach the proof this way, the key piece will be showing that the error due to using these approximations is sufficiently small. $\endgroup$ – amd May 3 at 0:53
  • $\begingroup$ @amd ... that's what I'm asking how to do. I'm fairly convinced the error bookkeeping can be done correctly with the structure of the proof above. Can I introduce a named function say $\rho(x) $ or something that bounds the error for both $f$ and $g$ simultaneously? $\endgroup$ – Gregory Nisbet May 3 at 1:13
  • $\begingroup$ See en.m.wikipedia.org/wiki/Chain_rule#Second_proof. $\endgroup$ – amd May 3 at 7:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.