How to expand $\partial^2(f\circ g)$ symbolically?

Question

I know that if $f,h\in C^\infty(F,\Bbb R)$ and $g\in C^\infty(E,F)$ for arbitrary real Banach spaces $E$ and $F$, then $\partial f^k\in C^\infty(F,\mathcal L^k(F,\Bbb R))$ and $g^k\in C^\infty(E,\mathcal L^k(E,F))$. Also I have the product rule (for real valued-valued functions) and chain rule that, it reads as

$$\partial(fh)=f\partial h+h\partial f,\quad \partial(f\circ g)=(\partial f\circ g)\partial g$$

where the product above means distribution of variables and composition, that is

$$\partial(fg)(x)=f(x)\cdot\partial h(x)+\partial g(x)\cdot f(x),\quad \partial(f\circ g)(x)=(\partial f\circ g)(x)\cdot\partial g(x)$$

where "$\cdot$" is composition of linear maps. However when I tried to express the second derivative of a composition symbollically, following the above product rule (for real-valued vector functions) and the chain rule, I got

$$\partial^2(f\circ g)=\partial((\partial f\circ g)\cdot\partial g)$$

but I dont know exactly how to interpret here the dot to know if I can continue using the chain rule to expand the expression further. If a plug a variable then

$$\partial^2(f\circ g)(x)\overset{?}{=}\partial\big((\partial f\circ g)(x)\partial g(x)\big)$$

but I dont know if the expression on the RHS makes sense.

My question is: there is a way to get a symbolic expression for these derivatives using the chain rule further? Some help will be appreciated, thank you.

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I know a generalized product rule that reads

$$\partial\varphi(f,g)=\varphi(\partial f,g)+\varphi(f,\partial g)$$ for any bilinear map $\varphi$, but I dont see how to apply it here.

Answer 1

Ok, I solved it... I didnt noticed that matrix multiplication is a bilinear map, lol, so I can apply the generalized product rule to it. Then we have that

$$\partial^2(f\circ g)(x)=\partial[(\partial f\circ g)(x)\cdot\partial g(x)]=\partial\varphi[(\partial f\circ g)(x),\partial g(x)]=\\=\varphi[(\partial^2 f\circ g)(x)\partial g(x),\partial g(x)]+\varphi[(\partial f\circ g)(x),\partial^2g(x)]=\\=\underbrace{(\partial^2 f\circ g)(x)}_{\in\mathcal L^2(F,\Bbb R)}\cdot\underbrace{[\partial g(x)]^2}_{\in\mathcal L(E, F)}+\underbrace{(\partial f\circ g)(x)}_{\in\mathcal L(F,\Bbb R)}\cdot\underbrace{\partial^2g(x)}_{\in\mathcal L^2(E,F)}$$

where $\varphi$ means "matrix multiplication" (assuming some matrix representation of the linear maps) and the above must be understood as a bilinear map from $E\to\Bbb R$ (with the power, as the dot, as composition of linear maps).

However, we must follow the order of differentiation to make meaningful how to plug values on the bilinear function, that is

$$\partial^2(f\circ g)(x)[v,h]=\partial(\partial(f\circ g)(x)v)h=\partial\big((\partial f\circ g)(x)\partial g(x)v\big)h=\\=\partial\big((\partial f\circ g)(x)D_v g(x)\big)h=(\partial^2 f\circ g)(x)[D_vg(x),D_hg(x)]+(\partial f\circ g)(x)D_hD_vg(x)$$

Answer 2

I would think about this for $f,g: \mathbb{R} -> \mathbb{R}$ first.

If $\partial ((f \circ g)(x)) = \partial (f'(g(x)) g'(x))$ Then let h = f' to make things easier to see and let's do this again $$ \partial(f'(g(x)) g'(x)) = \partial[h(g(x)) g'(x)] = \partial(h(g(x)) g'(x) + h(g(x)) g''(x) $$ where we have used product rule. Finishing with chain rule $$ \partial(f'(g(x)) g'(x)) = h'(g(x)) g'(x) g'(x) + h(g(x)) g''(x) $$ Substituting back in that $h = f'$ and therefore $h' = f''$ we have $$ \partial^2(f(g(x)) = \partial[f'(g(x)) g'(x)] = f''(g(x)) (g'(x))^2 + f'(g(x)) g''(x) $$ Hope this helps.