Composition of multivariate functions.

How can I understand a theorem about the composition of two functions:

$$g = \gamma: I \subset \mathbb{R} \rightarrow \mathbb{R}^m$$

$$f: \mathbb{R}^m \rightarrow \mathbb{R}$$

if it looks like this:

$$d(f \circ \gamma)(t_0) = df(\gamma(t_0))d\gamma(t_0)=\langle\nabla f(\gamma(t_0)), \gamma^{\prime}(t_0)\rangle$$

Isn't $$d(f \circ \gamma)(t_0)$$ just another way to write $$df(\gamma(t_0))$$? How can I understand this notation?

No. For a function $$f:\mathbb R^n \to \mathbb R^m$$, the differential takes in two inputs $$df:\mathbb R^n \times \mathbb R^n \to \mathbb R^m$$ It's linear in the second, but not normally linear in the first.

$$d(f\circ \gamma)(t_0)$$ Is the differential of $$f\circ \gamma$$ evaluated at $$t_0$$. One argument remains unevaluated, the result is a linear map $$\mathbb R\to\mathbb R^m$$.

$$df(\gamma(t_0))$$ Is the differential of $$f$$ evaluated at $$\gamma(t_0)$$. Again this is a linear map. But it takes in a vector. This vector would be the result of $$d\gamma(t_0)$$ applied to a real number, but this is not yet given. The final result $$d(f\circ \gamma)(t_0) d\gamma(t_0)$$ is a linear map $$\mathbb R \to \mathbb R^m$$.

The final equality is by a basic version of Riesz Representation - the linear map is the same as matrix multiplication with a certain row vector, and that vector is written out for you there.

PS in dimension one, $$df(x)(h)$$ is the multiplication of real numbers $$f'(x)h.$$ Here it's clear that the two terms you are asking about is different, and the stated equality is a generalisation of chain rule from highschool.

• I like your argument with the Riesz Representation :). That might be could for one of those question, where one argues with a way to powerful theorem for a "basic thing", don't you think so? Sep 21 '19 at 2:18
• @hal4math Yes, I name-dropped it in the hopes that it would help, what I really want is the line after which doesn't have a snappy slogan way of saying it, and one can hopefully ignore the Big Name if it's not familiar to them... Sep 21 '19 at 2:22
• @CalvinKhor Later in the script, $\gamma (t_0) = x_o, \gamma^{\prime}=v_0$ are definded to justify: $\partial_{v_0} f(x_0) = (f \circ \gamma)^{\prime}(t_0) = \langle\nabla f(x_0), v_0\rangle$. Of course I understand the second and third part, but could you explain what the first part means? Why do we take the partial with respect to the derivative?
– MJP
Sep 21 '19 at 2:50
• @Markus sorry I'm away from keyboard but that looks like a directional derivative, does that help? Sep 21 '19 at 3:16
• Oh, how could I not see. Thanks, that explains it (since taking the directional derivative takes sum of partials with components of the vector in which direction we want to derive as factors, which is just like the inner product on the right).
– MJP
Sep 21 '19 at 3:30

Now, I think I get where maybe your confusions lies. As pointed out in the comment above, more brackets might be helpful.

So, since $$(f \circ \gamma)(t_0) = (f(\gamma(t_0))$$ is a composition of two functions, when taking the derivative you need the chain rule. Now you need to be aware of what each $$d$$ refers two. In the expression $$d(f\circ \gamma)(t_0) = d(f(\gamma(t_0)))$$ the differential is still to be taking with the chain rule which gives $$d(f \circ \gamma)(t_0) = (df)(\gamma (t_0)) \circ (d\gamma)(t_0),$$ where now the differentials (or total derivatives) are only with respect to one function. So $$(df)(\gamma(t_0)$$ is to read as, the differential $$df$$ evaluated at point $$\gamma(t_0)$$.

Now, since we are in $$\mathbb{R}$$ or $$\mathbb{R}^n$$ respectively, (and everything is smooth enough, e.g. the total derivatives exists, I assume) you can identify the differential with the jacobian matrix, where the composition $$\circ$$ becomes a Matrix Matrix multiplication - which in your case you expressed it with a scalar product.