A proof of the product rule using the single variable chain rule?

Credit is due to this video by Mathsaurus, but also requires the sum and power rules for derivatives.

Let $u$, $v$ be appropriate functions. Consider $f = (u+v)^2$. By the chain rule, $$f^{\prime} = 2(u+v)(u^{\prime}+v^{\prime})=2(uu^{\prime}+uv^{\prime}+vu^{\prime}+vv^{\prime})\text{.}$$ Now, expand $f$ to obtain $f = u^2+2uv+v^2$, and then we have $$f^{\prime}=2uu^{\prime}+2(uv)^{\prime}+2vv^{\prime}=2[uu^{\prime}+(uv)^{\prime}+vv^{\prime}]\text{.}$$ It follows immediately that $$(uv)^{\prime}=uv^{\prime}+vu^{\prime}\text{.}$$

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, $$2ff’+2fg’+2f’g+2g’g=2ff’+2(fg)’+2gg’$$ $$2fg’+2f’g=2(fg)’$$ $$(fg)’=fg’+f’g$$

It all depends on what other principles you have available to you, but since you mention logarithmic differentiation, note that $\log(fg) = \log(f) + \log(g)$. Take the derivative of both sides and multiply by $fg$, to get $(fg)' = f'g + fg'$.