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I have often read that it is less precise to state the chain rule using Leibniz's notation as opposed to Lagrange's notation. I don't understand this claim because it seems to me that the two statements are identical, and, if anything, Leibniz's formulation is 'cleaner'. If $y=f(u)$, and $u=g(x)$ (and $f$ is differentiable at $u$, and $g$ differentiable at $x$), then

$$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx} $$

Equally, we could write

$$ (f \circ g)'(x) = (f' \circ g)(x)g'(x) $$

It seems simple to show that these two statements mean the same thing:

$$ \frac{dy}{dx}\equiv(f \circ g)'(x) $$

because they both refer to the change in $y$ (where $y=(f \circ g)(x)$) divided by the change in $x$. More formally, they both refer to the same limit expression: $$ \lim_{\Delta x \to 0}\frac{f(g(x+\Delta x))-f(g(x))}{\Delta x} $$ We can also verify that $\frac{dy}{du}\equiv(f' \circ g)(x)$ and that $\frac{du}{dx} \equiv g'(x)$. However, the fact that the two notations refer to the same statement does not mean that we can't prefer one over the other. So what dangers are there in using Leibniz's notation for the chain rule? I have always found it to be simpler and more intuitive, but perhaps there are things which make Leibniz's notation misleading.

Joe
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  • $(f\circ g)'(x)$ means $\lim_{\Delta x\to 0}\dfrac{f(g(x+ \Delta x)) - f(g(x))}{\Delta x}$ – peek-a-boo Aug 20 '20 at 21:13
  • @peek-a-boo Thanks for correcting my error. I've fixed it now – Joe Aug 20 '20 at 21:14
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    Once you have done enough analysis, there are no "dangers", one uses either of the two notations, depending on the problem. Of course the $f \circ g$ notation is the reference in case of doubt... mainly for a student, but the most handy is of course Leibniz's. – Jean Marie Aug 20 '20 at 21:16
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    The $y$'s on the two sides of the equation have different meanings, and this difference isn't made clear in the notation. If you know what you're doing there's of course there's no danger, but the thing is Leibniz's notation constantly requires the reader to do a lot of "interpretation", something which is not needed for the primed notation. I'm not sure how the notation $\dfrac{dy}{du}$ automatically suggests the composition $f'\circ g$. In this case, how would you write down $f'$ in Leibniz's notation? I just wrote an answer a while back. – peek-a-boo Aug 20 '20 at 21:18
  • @Jean Marie Thank you for responding. Why is the $f\circ g$ notation a good one to fall back on? – Joe Aug 20 '20 at 21:23
  • @peek-a-boo Thanks, that was quite a useful answer. I think my main source of confusion at this point is that I don't quite understand how the $y$'s are different. Also, I thought $dy/du$ does suggest $(f' \circ g)(x)$, as $(f' \circ g)(x)=f'(g(x))=f'(u)=\text{Change in $y$ with respect to $u$}$. – Joe Aug 20 '20 at 21:34
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    How would you write the abstract functions $f', f'\circ g$ and $(f\circ g)'$ in terms of Leibniz notation? – peek-a-boo Aug 20 '20 at 21:37
  • related – Jackozee Hakkiuz Aug 21 '20 at 00:54
  • To answer your question, $f \circ g$ is the "reference" because in this way, the relationship between $y$ and $u$, on one side and $u$ and $x$ on the other side is explicitly attached to $f$ or $g$ for which we can say for example they are $C^1$ but not $C^2$, something that we cannot "relate" with Leibniz notations. – Jean Marie Aug 21 '20 at 06:16
  • One should perhaps as well have a look through nonstandard analysis "glasses" (I hardly know the field). – Jean Marie Aug 21 '20 at 08:27

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It is possibly because beginners may think there is "cancellation" in $$ \frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}\; $$ while there is really no.

As long as interpreted correctly, both sets of notations are "precise".

See also excellent discussions of notations in this post:

https://mathoverflow.net/a/366118/164038