# Compute the derivative of the function $g \circ (f_1,\ldots,f_m)$

Inspired by this quesion, I'm trying to compute the derivative of the below function. Could you please verify if my proof looks fine or contains logical gaps/errors? Thank you so much for your help!

Let $$X$$ be a metric space and $$F$$ a normed vector space. Suppose $$f_j: X \to \mathbb R$$ is differentiable at $$a$$ for all $$j = \overline{1,n}$$ and $$g: \mathbb R^n \to F$$ is differentiable at $$(f_1(a),\ldots, f_n(a))$$. Prove that $$g \circ (f_1,\ldots,f_m)$$ is differentiable at $$a$$ and compute its derivative.

My attempt:

Lemma: Assume that $$f_j:X \to E_j$$ is differentiable at $$a$$ for all $$j = \overline{1,n}$$. Then $$\begin {array}{l|rcl} f & X & \longrightarrow & E_{1} \times \cdots \times E_{n} \\ & x & \longmapsto & (f_1 (x), \ldots, f_n(x)) \end{array}$$ is differentiable at $$a$$ and $$\partial f(a) = (\partial f_1 (a), \ldots, \partial f_n(a))$$.

Let $$f = (f_1,\ldots,f_m)$$. Then $$g \circ (f_1,\ldots,f_m) = g \circ f$$. It follows from our Lemma that $$f$$ is differentiable at $$a$$. By the chain rule and our Lemma, we get \begin{aligned} \partial (g \circ f) (a) &= \partial g (f(a) \circ \partial f (a) \\ &= \sum_{j=1}^n \partial_j g(f(a)) \cdot \partial f_j (a)\end{aligned}

Update:

Let $$\{e_j \mid 1 \le j \le n\}$$ be the standard basis of $$\mathbb R^n$$.

Because $$g \circ f \in F^X$$, $$\partial (g \circ f)(a) \in \mathcal L(X,F)$$. In other words,

$$\begin {array}{l|rcl} \partial (g \circ f)(a) & X & \longrightarrow & F \\ & x & \longmapsto & \partial (g \circ f)(a)(x) \end{array}$$

Similarly, we have

$$\begin {array}{l|rcl} \partial f_j(a) & X & \longrightarrow & \mathbb R \\ & x & \longmapsto & \partial f_j(a)(x) \end{array} \quad \text{and} \quad\begin {array}{l|rcl} \partial g(f(a)) & \mathbb R^n & \longrightarrow & F \\ & v & \longmapsto & \partial g(f(a))(v) \end{array}$$

We have $$\partial g (f(a) \circ \partial f (a)$$ is a continuous linear map such that

$$\begin {array}{l|rcl} \partial g (f(a) \circ \partial f (a) & X & \longrightarrow & \mathbb R \\ & x & \longmapsto & \left (\sum_{j=1}^n \partial_j g(f(a)) \cdot \partial f_j (a) \right )(x) = \sum_{j=1}^n \partial_j g(f(a)) \cdot \partial f_j (a) (x)\end{array}$$

Here $$\partial f_j (a)(x) \in \mathbb R$$ and $$\partial_j g(f(a)) = \partial g(f(a)) (e_j) \in F$$. Hence $$\partial_j g(f(a)) \cdot \partial f_j (a) (x) \in F$$ and thus $$\sum_{j=1}^n \partial_j g(f(a)) \cdot \partial f_j (a) (x) \in F$$.

The derivative of $$g\circ f$$ at $$a$$ should be $$\sum_{j=1}^n \partial_j g(f(a)) \cdot \partial f_j$$.
Notice that the function $$g\circ f$$ maps $$X$$ to $$F$$ so, for instance, if $$F=\mathbb{R}$$ and $$n>1$$ then the derivative of $$g\circ f$$ at $$a$$ is a real number whereas $$\sum_j\partial_kg(f(a))\cdot\partial f_j(a)$$ is an $$n$$-dimensional vector.
• Hi, from my understanding: Because $\partial (g \circ f)(a) \in \mathcal L(X,F)$, we have $\partial (g \circ f)(a) \in \mathcal L(X,F)$. Hence the derivative of $g\circ f$ at $a$ is a map, not just a number. I have updated my post with details of my thinking. Could you please have a look at it? Nov 9, 2019 at 13:48
• $\partial(g\circ f)$ is a map from $X$ to $F$. $\partial(g\circ f)(a)$ is an element of $F$. Nov 10, 2019 at 7:46