Let $$A(x)=sin(x), A'(x)=cos(x)$$ $$B(x)=e^x, B'(x)=e^x$$ $$C(x)=x^2+x,C'(x)=2x+1$$ and let $$f(x) = A(B(C(x))) =sin(e^{x^2+x})$$
In Lagrange's notation we express the chain rule as follows: $$ f'(x)=A'(B(C(x))) \cdot B'(C(x)) \cdot C'(x)=cos(e^{x^2+x})\cdot e^{x^2 +x} \cdot (2x+1)$$
It's intuitively obvious to me what the meaning of each multiplicands stands for. It's simply a function composition. We take the derivative of the outermost function and use the input of the original function as the input of the derivative. Then we "peel off" the outer function and repeat.
In Leibniz's notation we have $$\frac{d}{dx}f(x)= \frac{dA}{dB} \cdot \frac{dB}{dC} \cdot \frac{dC}{dx}$$
Do these 3 multiplicands correspond in order to the multiplicands in Lagrange's notation? I think that my problem with this notation is that the term $\frac{dA}{dB}$ does not contain the information that the input of the derivative of $A$ is actually $B(C(x))$ in this case not just $B(x)$ as someone might think when seeing the multiplicand without context.
On the other hand, in Lagrange's notation each multiplicand has complete information about its input. Am I understanding this correctly (this information is represented in the notation implicitly so you have to look at all the multiplicands first)? In other words, were I to compute $f'(x)$ in Lagrange's notation you can give me the multiplicands separately and I would compute them independently but in Leibniz's I can't do that since $\frac{dA}{dB}$ has no meaning without context?
If possible I'd like to hear is an intuitive description of what the sequence of multiplicands means that I can internalize and use whenever I see the chain rule in Leibniz's notation or maybe what its advantages are.