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I've been asked to proof the product rule. I am fine with that, my proof was accepted completely. Nevertheless, for me personally the very last step seems to be not totally incorrect, but, let say, "intuitive" or "inductive" to some extend. Not to confuse people, will mention few fateful points of my reasoning:

At some moment I've proved that: $$f(x + \delta) = A + B \times \delta + o(E(\delta))$$ where:

  1. $o(E(\delta))$ means "of less order than $\delta$", i.e. $\lim_{\delta \to 0}\frac{E(\delta)}{\delta} = 0$;
  2. $A = f(x)$;
  3. $B = f'(x)$.

After that I took both $f(x + \delta)$ and $g(x + \delta)$ assumed to be differentiable at the point $x$. Hence: $$f(x + \delta) \times g(x + \delta) = [f(x) + f'(x)\delta + o(E(\delta))] \times [g(x) + g'(x)\delta + o(E(\delta))]$$ $$= f(x)g(x) + [f(x)g'(x) + g(x)f'(x)] \times \delta + \biggl( [f(x) + f'(x)\delta]o(E(\delta)) + [g(x) + g'(x)\delta]o(E(\delta)) + [o(E(\delta))]^2\biggr)$$


HERE GOES MY QUESTION:

I've noticed that the last equation could be reformulated as: $$A = f(x)g(x)$$ $$B = f(x)g'(x) + g(x)f'(x)$$ $$o(E(\delta)) = \biggl( [f(x) + f'(x)\delta]o(E(\delta)) + [g(x) + g'(x)\delta]o(E(\delta)) + [o(E(\delta))]^2\biggr)$$

The question is: shouldn't I proof somehow (have no idea how, actually) that the pattern above is applicable here? Last step seems to be more of blinded guessing rather than an undeniable logical deduction...

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  • $\begingroup$ I don't think you ever assumed $f(x+\delta)$ or $g(x+\delta)$ to be differentiable at $x$. $\endgroup$
    – Kenny Lau
    Dec 25, 2017 at 0:34
  • $\begingroup$ @KennyLau why do you think so? $\endgroup$
    – Zazaeil
    Dec 25, 2017 at 8:34

1 Answer 1

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I prefer to write $$ f(x+\delta)=f(x)+\delta f'(x)+\delta\varphi(\delta) $$ where the condition $\lim_{\delta\to0}\varphi(\delta)=0$ is equivalent to $f$ being differentiable at $x$, with derivative $f'(x)$. More precisely: write, for $\delta\ne0$, $$ \varphi(\delta)=\frac{f(x+\delta)-f(x)}{\delta}-a $$ Then $f$ is differentiable at $x$, with derivative $a$ if and only if $\lim_{\delta\to0}\varphi(\delta)=0$.

It's not so different from your notation, actually, but less confusing, in my opinion.

Suppose the same for $g$, so $$ g(x+\delta)=g(x)+\delta g'(x)+\delta\psi(\delta) $$ and $\lim_{\delta\to0}\psi(\delta)=0$.

Now we have \begin{align} f(x+\delta)g(x+\delta) &=\bigl(f(x)+\delta f'(x)+\delta\varphi(\delta)\bigr) \bigl(g(x)+\delta g'(x)+\delta\psi(\delta)\bigr) \\ &=f(x)g(x)+\delta\bigl(f'(x)g(x)+f(x)g'(x)\bigr) \\ &+\delta\bigl( \begin{aligned}[t] &\delta f'(x)g'(x)+ \varphi(\delta)g(x)+ \delta f'(x)\psi(\delta)+{}\\ &\delta\varphi(\delta)g'(x)+ f(x)\psi(\delta)+\delta\varphi(\delta)\psi(\delta)\bigr) \end{aligned} \\ &=f(x)g(x)+\delta\bigl(f'(x)g(x)+f(x)g'(x)\bigr) + \delta\eta(\delta) \end{align} and $\lim_{\delta\to0}\eta(\delta)=0$, proving the statement.

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  • $\begingroup$ Actually, your proof is exactly the same besides the fact that I prefer be slightly more explicit by marking $o(...)$ what you reference as $\eta(\delta)$. The essence remains to be the same, or, at least, I don't see any change. So the question still holds: who gave you a right to apply the pattern from above without proving that it is indeed reasonable? $\endgroup$
    – Zazaeil
    Dec 25, 2017 at 8:37
  • $\begingroup$ @СергейБоголюбов The first paragraph is the key: the function $\varphi$ is well defined for $\delta\ne0$ and its limit being $0$ is equivalent to differentiability of $f$ at $x$ $\endgroup$
    – egreg
    Dec 25, 2017 at 10:19
  • $\begingroup$ The fact that $\delta \times \eta(\delta)$ and $\lim_{\delta \to 0}\eta(\delta)$ actually means that $lim_{\delta \to 0} \frac{\delta \times \eta(\delta)}{\delta} = 0$. In other words, first paragraph does not make difference. $\endgroup$
    – Zazaeil
    Dec 25, 2017 at 11:59
  • $\begingroup$ @СергейБоголюбов Sorry, but I don't understand. Since $\eta(\delta)=\frac{f(x+\delta)g(x+\delta)-f(x)g(x)}{\delta}-(f'(x)g(x)+f(x)g'(x))$, the fact that $\lim_{\delta\to0}\eta(\delta)=0$ (which is true by simple computation from $\varphi$ and $\psi$) implies that $fg$ is differentiable at $x$ with derivative $f'(x)g(x)+f(x)g'(x)$. $\endgroup$
    – egreg
    Dec 25, 2017 at 13:54

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