Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.

Consider two functions $f,g$ and the expression $\frac{d}{dx} (f+g)^2$. Directly applying the chain rule, this becomes 2(f+g)(f’+g’).

Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $\frac{d}{dx}(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.

Since these two expressions are equal, we have $$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$ Using some algebra, $$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$ $$2fg’+2f’g=2(fg)’$$ $$(fg)’=fg’+f’g$$

Here’s a relatively simple one that only uses the single variable chain rule and some elementary algebra.

Consider two functions $f,g$ and the expression $\frac{d}{dx} (f+g)^2$. Directly applying the chain rule, this becomes $2(f+g)(f’+g’)$.

Instead of directly using the chain rule, one can also multiply out $(f+g)^2$, and the expression becomes $\frac{d}{dx}(f^2+2fg+g^2)$, and we can differentiate term by term to obtain $2ff’+2(fg)’+2gg’$.

Since these two expressions are equal, we have $$ 2(f+g)(f’+g’)=2ff’+2(fg)’+2gg’$$ Using some algebra, $$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$ $$2fg’+2f’g=2(fg)’$$ $$(fg)’=fg’+f’g$$