Using chain rule to differentiate $f(x)=a(x)b(x)$? Why can I not apply the chain rule to a product in the following way.
If we have some product:
$$f(x)=a(x)b(x)$$
Consider the multiplication of b by a as another’s function so that:
$$f(b(x))=ab$$
So that 
$\frac{df}{dx} = f’(b)b’(x)$
Something feels very wrong. But I can’t put my finger on it.

If you can find such a composition $A\circ b = a(x)b(x)$, then the chain rule gives you the product rule. You can find $A\circ b$ just if b is invertible.


*

*Because $a(x)$ and $b(x)$ are both functions of $x$, you might not be able to find a "multiplication function" $A$ that lets you write the product  $a(x)b(x)$ as a chain rule composition $A(b(x))$. 
This is because $A$ only sees the output of the function $b(x)$. It doesn't see what $x$ is. If it can't see what $x$ is (or figure it out based on the value of $b(x)$), then it can't compute $a(x)$ and so it can't compute $a(x)b(x)$.

*When you can recover the value of $x$ from the value of $b(x)$, we say that $b$ is invertible.

*If $b$ is invertible, then it has an inverse $b^{-1}$ which recovers the value of $x$. So, if $y=b(x)$ then $b^{-1}(y) = x$. 
If $b$ is invertible, then you can define the function $A(y) \equiv a(b^{-1}(y)) \cdot y$. 

*Then notice that by definition, $A(b(x)) = a(b^{-1}(b(x))) \cdot b(x) = a(x)b(x)$, exactly as we want it. We have written $a(x)b(x)$ as a chain rule composition $A(b(x))$.

*When we take the derivative using the chain rule, we find that $\frac{d}{dx}A(b(x)) = a^\prime(x)b(x) + a(x)b^\prime(x)$, exactly as you want. Though because the definition of $A$ depends on a product, you either have to use the product rule anyways to differentiate it, or use the limit definition of the derivative to compute it.

In multivariable calculus, the chain rule directly gives you the product rule.
What you might be looking for, after all, is a way to use the chain rule as a fundamental tool for proving all of the other rules such as the product rule. You can do this with multivariable calculus.


*

*The multiplication function is $M(u,v) = uv$. Its derivative is $DM(u,v) = \langle v, u\rangle$.

*If your two functions are $a(x)$ and $b(x)$, then their product is $g(x) = a(x)b(x)$. 

*We can write $g$ as a composition: $g(x) = M(a(x), b(x))$. 

*The derivative of $g$ is therefore the derivative of $M(a(x), b(x))$---and we can compute this latter derivative using the chain rule.

*By the chain rule, 
\begin{eqnarray*}
D\left[M(a(x),b(x))\right] &= DM( a(x), b(x))\bullet D\langle a(x), b(x)\rangle\\
&= \langle b(x), a(x)\rangle \bullet \langle a^\prime(x), b^\prime(x)\rangle\\
&= b(x)a^\prime(x) + a(x)b^\prime(x)
\end{eqnarray*}
