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: 


*

*$o(E(\delta))$ means "of less order than $\delta$", i.e.
$\lim_{\delta \to 0}\frac{E(\delta)}{\delta} = 0$;

*$A = f(x)$;

*$B = f'(x)$.


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...
 A: 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.
