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:
- nLab – Hadamard lemma
- Second answer in this m.se tread
- 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)?
Best Answer
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 $\frac{df}{dt}(g(t))$.