# Use Chain rule rigorously with correct notation.

He gave the Chain rule as follows:

Then he proved the following example:

What confused me is the second equality in the above Example 2.4.5, which is underlined by a yellow line. What exactly is this?

Here is my understanding: I understand that it is a "compact" style of writing a function. I mean, for example, given any point $$a$$ in $$\mathbf{R}$$. The author tries to convey that (using the chain rule)$$D(g \circ f)(a)= (D_1g(f(a)), \cdots, D_ng(f(a)) \circ \left( \begin{matrix} Df_1(a) \\ \vdots \\ Df_n(a) \end{matrix} \right)$$. Then we can use legitimate matrix multiplication to get the final result.

I think it's not rigorous:

1. He writes $$(Dg\circ f)$$ as $$((D_1g,\cdots,D_ng)\circ f)$$. However, each $$D_ig$$ is a function from $$\mathbf{R}^n$$ to $$\mathbf{R}$$ and he just wrote them in a row vector, pretending they have accepted $$f$$'s output as input and have output a scalar in $$\mathbf{R}$$. I've never seen expressions like this before. Besides, how to prove $$Dg=(D_1g,\cdots,D_ng)$$? Is the total derivative equal to some undefined list of partial derivatives?

2. The same expression appeared in $$\left( \begin{matrix} Df_1 \\ \vdots \\ Df_n \end{matrix} \right)$$

3. I think if someone doesn't know in advance what the author tries to convey. These are all invalid expressions. So how should a beginner like me use these expressions correctly to prove more complicated statements?

• @Troposphere Yes, you are right. But I am emphasizing $(Dg) \circ f$, not $D(g \circ f)$, I mean $D(g \circ f) = ((Dg) \circ f) \circ Df$, see theorem 2.4.1 Commented Jul 31, 2021 at 4:06
• Sorry, I misread the actual quote in the question. Commented Jul 31, 2021 at 12:04

The author implicitly identified derivative at a point $$a$$ as its matrix representation.

In fact, in example 2.4.5, the author emphasized this identification:

the derivative $$D(g \circ f): \mathbf{R} \rightarrow End(\mathbf{R}) \simeq \mathbf{R}$$

So if it was me, I would write the equation in Example 2.4.5 as follows:

for any $$a \in \mathbf{R}$$,

\begin{align} [D(g \circ f)(a)]_{1 \times 1} &= [Dg(f(a))]_{1 \times n} \circ [Df(a)]_{n \times 1}\tag{Jacobi matrix notation} \\ &=[D_1g(f(a)), \cdots, D_ng(f(a))]_{1 \times n} \circ \left( \begin{matrix} Df_1(a) \\ \vdots \\ Df_n(a) \end{matrix} \right)_{n \times 1} \\ &= \sum_{1\leq i\leq n} (D_ig(f(a))) \cdot Df_i(a) \end{align}

I don't think the "compact style" of writing an equation is good for learning.