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Consider $f:\mathbb{R}^n\rightarrow\mathbb{R}^m$ and $g:\mathbb{R}^m\rightarrow\mathbb{R}^k$. Then $(g\circ f):\mathbb{R}^n\rightarrow\mathbb{R}^k$ and, if both of them are differentiable, $[D(g\circ f)_p]=(Dg)_{f(p)}\cdot (Df)_p$. If these functions are two times differentiable, then

$D(D(g\circ f))_p=(D^2g)_{f(p)}\cdot (Df)^2_p + (Dg)_{f(p)}\cdot (D^2 f)_p$.

I'm trying to figure out what $(Df)^2_p$ means. Since it is a $m\times n$ matrix I cannot multiply them. Can someone help me?

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    It's a tensor. To get an idea of what's going on: say we have $f(x,y,z)$ (scalar valued). Then $Df=\nabla f = \langle f_x,f_y,f_z \rangle$ (a vector of functions). $D^2f=H$ (the Hessian matrix) which is a $3 \times 3$ matrix filled with second partials. $D^3f$ is a $3 \times 3 \times 3$ cube filled with third partials. etc. So if your first derivative $Df$ is an $m \times n$ matrix, $D^2f$ is a vector of matrices (or better to start using tensors).2012-04-08
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    Are you sure? Because what I have in my equation is $(Df)_p\cdot (Df)_p$, not $(D^2 f)_p$. I know what $(D^2 f)_p$ is, but this product is what is taking my sleep away at night :)2012-04-08
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    Yes. I'm sure. To make sense out of "multiplying" these things using matrices and such, you'll need to pull apart $D(f)$ into a big column vector so that $D^2(f)$ is again a matrix. But in the end this is not a great way to handle such matters.2012-04-08
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    @Bill Gustavo is asking about $(Df)_p^2$, not $(D^2f)_p$.2012-04-08

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