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

A while back I saw someone claim that you could prove the product rule in calculus with the single variable chain rule. He provided a proof, but it was utterly incomprehensible. It is easy to prove from the multi variable chain rule, or from logarithmic differentiation, or even from first principles. Is there an actual proof using just the single variable chain rule?

Solution 1:

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 $$(uv)^{\prime}=uv^{\prime}+vu^{\prime}\text{.}$$

Solution 2:

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

Solution 3:

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 $(fg)' = f'g + fg'$.