# Product rule proof. Derivatives.

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...

• I don't think you ever assumed $f(x+\delta)$ or $g(x+\delta)$ to be differentiable at $x$. – Kenny Lau Dec 25 '17 at 0:34
• @KennyLau why do you think so? – Sereja Bogolubov Dec 25 '17 at 8:34

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 $\lim_{\delta\to0}\varphi(\delta)=0$.
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.
• 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? – Sereja Bogolubov Dec 25 '17 at 8:37
• @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 $x$ – egreg Dec 25 '17 at 10:19
• 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. – Sereja Bogolubov Dec 25 '17 at 11:59
• @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 $f'(x)g(x)+f(x)g'(x)$. – egreg Dec 25 '17 at 13:54