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Suppose that $f \in C^\infty(\mathbb{R}^n,\mathbb{R})$ and $y = Ax + b$ is an affine transformation of $\mathbb{R}^n$. Does there exist a simple formula for the (higher order) partial derivatives of $f(y)$ with respect to $x$?

My main insight is that any second order partial derivative of $y$ will be zero, but I get lost in my calculations.

mixotrov
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1 Answers1

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Let $h(x):=Ax+b$ and $g:=f\circ h,$ i.e. $$g(x)=f(Ax+b).$$ Then, $Dh_x=A$ and by the chain rule, $Dg_x=Df_{Ax+b}\circ A,$ i.e. $$Dg_x(u)=Df_{Ax+b}(Au).$$ Iterating this, you get $$D^ng_x(u_1,\dots,u_n)=D^nf_{Ax+b}(Au_1,\dots,Au_n). $$

Anne Bauval
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  • Unfortunately, I'm not that familiar with total derivatives. Why do we get $n$ arguments when iterating? My first guess would $D^ng_x(u)=D^nf_{Ax+b}(A^nu)$. And how would I extract a particular partial derivative, say $\dfrac{\partial^3}{\partial{x_1}\partial^2{x_2}}$? – mixotrov Jan 26 '23 at 15:32
  • These are total differentials. The first differential $Dg$ at some point $x$ is a linear map, which we apply to a vector $u.$ The partial derivatives are related to it by: $Dg_x(u)=\sum\frac{\partial g}{\partial x_i}(x)u_i$ or equivalently: $\frac{\partial g}{\partial x_i}(x)=Dg_x(e_i)$ where $e_i$ is the $i$-th vector of the canonical basis. The $n$-th differential $D^ng$ at $x$ is an $n$-multilinear map (symmetric, by Schwarz theorem). $\dfrac{\partial^3g}{\partial{x_1}\partial^2{x_2}}(x)=D^3g_x(e_1,e_2,e_2).$ – Anne Bauval Jan 26 '23 at 15:49