In differentiation can you prove the product rule using the sum rule by breaking apart the product

Question

for example if you let $f(x)=g(x)h(x)$
then $f(x)$ can be written as $f(x)=g(x)+g(x)+g(x)...$ appearing $h(x)$ amount of times

therefore using the sum rule, $f'(x)$ can be written as $f'(x)=g'(x)+g'(x)+g'(x)...$
where $g'(x)$ appears $h(x)$ amount of times, therefore: $f'(x)=g'(x)h(x)$

same applies for the other way of $f'(x)=h'(x)+h'(x)+h'(x)...$ so $f'(x)=h'(x)g(x)$

but the product rule is $f'(x)=g'(x)h(x)+g(x)h'(x)$

therefore how can this be shown using this reasoning with the sum rule, or where have I gone wrong in my reasoning so far, any help would be greatly appreciated.