# Replacing function with linear approximation inside proof of chain rule

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:

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

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

• 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. – amd May 3 at 0:53
• @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? – Gregory Nisbet May 3 at 1:13
• – amd May 3 at 7:28