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Since I've started reading up on differential geometry I keep coming across something that's bothering me, which is the chain rule. The standard chain rule from calculus is

$$ \frac{df}{dt}(g(t)) = f'(g(t))\cdot \frac{dg}{dt}(t), $$ which I could also write as $$ \frac{d f}{dt}(g(t)) = \frac{\partial f}{\partial g(t)}(g(t))\cdot \frac{dg}{dt}(t). \quad \quad (1) $$

Now when I'm reading these differential geometry texts I keep seeing the following strange approach to the chain rule (I'll stay in 1 dimension for simplicity). The function $g$ is specified explicitly and in the simplest case it could be $g(t) = tx$. Then the chain rule is always given as

$$ \begin{align} \frac{d f}{dt}(tx) = \frac{\partial f}{\partial x}(tx) \frac{d (tx)}{dt}. \end{align} $$

Replacing $tx$ by $g(t)$ (so we can compare it with the usual chain rule above) we have $$ \begin{align} \frac{d}{dt}f(g(t)) = \frac{\partial f}{\partial \color{red}{ \textbf{x}}}(g(t)) \cdot \frac{dg}{dt}(t), \quad \quad (2) \end{align} $$ where I have highlighted the problematic issue. Why have we $x$ in the denominator here instead of $g(t)$ as in the standard chain rule in (1) above?

Some places, among many others, where I have seen this are:

  1. nLab - Hadamard lemma
  2. Second answer in this m.se tread
  3. The book Introduction to Manifolds by Loring Tu.

So what is going on, is this some 'convention' in which this $x$ actually refers to $g(t) = tx$ or have I misinterpreted something and does the chain rule in (2) somehow agree with the standard chain rule (1)?

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    $\begingroup$ The left-hand side should be $(f \circ g)'(t)$ in the "standard chain rule", not $\frac{df}{dt}(g(t)) = f'(g(t))$. $\endgroup$ – Alex Provost Jul 17 '18 at 17:44
  • $\begingroup$ You shouldn't write $\frac{\partial f}{\partial \color{red}{g(t)}}(g(t))$ because $g(t)$ is not a variable. $\endgroup$ – Luca Bressan Jul 17 '18 at 17:49
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    $\begingroup$ And in the references you cite, $f$ is a multivariate function, so $\frac{\partial f}{\partial x^i}(tx)$ refers to the $i$th partial derivative of $f$ evaluated at the vector $tx$. In the 1-dimensional case you highlight, this is just the usual derivative $f'(tx) = \frac{df}{dx}(tx)$. I actually don't agree with your interpretation of the standard chain rule; you shouldn't write $\frac{\partial f}{\partial g(t)}$. $\endgroup$ – Alex Provost Jul 17 '18 at 17:49
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    $\begingroup$ @eurocoder The Leibniz notation relies on specific "variables". In your case, the "variable" for the domain of $g$ if $t$, and the "variable" for the domain of $f$ is $x$. So I still disagree with your notation (and use of partials instead of $d$). $\endgroup$ – Alex Provost Jul 17 '18 at 18:28
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    $\begingroup$ @eurocoder The second expression in your chain of equalities has no meaning; the first and third agree; and the fourth means $f'(g(t))$, which is not the same as $f'(g(t))g'(t)$. $\endgroup$ – Alex Provost Jul 17 '18 at 18:31
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I think we cleared most things up in the comments: in Leibniz notation, and to be very precise, one should write something like

$$ \frac{d(f \circ g)}{dt}(t) = \frac{df}{dx}(g(t)) \frac{dg}{dt}(t).$$

One will often see shorthand forms like $\frac{d}{dt}f(g(t))$ for the left-hand side, but that should never be written as $\frac{df}{dt}(g(t))$.

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  • $\begingroup$ Oh and just to mention something else Alex cleared up which might help someone in future - in differential geometry notation is abused such that, in the same expression, the notation $x^i$ may stand for both the coordinates of a particular point $x$ and also the components of some totally unrelated dependent variable used in the denominater of a partial derivative! $\endgroup$ – eurocoder Jul 17 '18 at 19:53

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