Proof of product rule using little-o definition of derivative.

Question

The derivative is most commonly defined as a limit of a difference quotient, $$f'(x) := \lim_{h\to 0}\frac{f(x + h) - f(x)}{h}.$$ However, the derivative can also be characterized as the unique number $f'(x)$ such that $$f(x + h) = f(x) + f'(x)h + o(h)$$ holds. I was wondering about how to prove the product rule with this definition. Let $f, g:\mathbf R\to\mathbf R$. May we then compute the following? $$\begin{align*} f(x+h)\cdot g(x + h) &= (f(x) + f'(x)h + o(h))\cdot(g(x) + g'(x)h + o(h))\\ &= f(x)g(x) + (f'(x)g(x) + g'(x)f(x))h + f'(x)g'(x)h^2\\ &+ o(h)(f(x) + f'(x)h + g(x) + g'(x)h) + o(h)o(h)\\ &= f(x)g(x) + (f'(x)g(x) + g'(x)f(x))h + f'(x)g'(x)h^2 + o(h) \end{align*} $$ And if so, would this computation reduce the proof of the product rule to showing that $f'(x)g'(x)h^2$ is $o(h)$?