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The derivative of $f:\mathbb{R}^N \to \mathbb{R}^M$ is of the form $f':\mathbb{R}^N \to \mathbb{R}^{M \times N}$. I'd like to know how the double derivative look like, i.e, how would $f''$ be ? It maps from $\mathbb{R}^N$ to which space?

PS : Please suggest some good references on this topic.

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    Well... applying what you wrote, to $\mathbb R^{M\times N\times N}$.2012-06-12
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    @did : An $M\times N\times N$ matrix to be defined as a linear map we need to define matrix multiplication for matrices of the form $L\times M\times N$. Where can I find this stuff? I wonder which book deals with this kind of stuff?2012-06-12
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    The main stumbling block to seeing how to construct the double derivative from the first might be an insufficient altitude of abstraction. The derivative of $f:V\to W$ is $f\,':V\to\hom(V,W)$ as the answers state, but $\hom(\Bbb R^n,\Bbb R^m)\cong\Bbb R^{m\times n}$ so we can rewrite the derivative as a matrix function. Similarly $\hom(V,\hom(V,W))\cong \hom(V\otimes V,W)$ and $\Bbb R^n\otimes\Bbb R^n\cong \Bbb R^{n\times n}$ (these are tensor products) so the double derivative can be written as $\Bbb R^m\to\Bbb R^{m\times n\times n}$.2012-06-12
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    The derivative of $f$ is a linear map with an $M\times N$ matrix. So the second derivative will be a linear map with an $(M\times N)\times N$ matrix. The thing in the brackets is just "some number". Also you don't need to multiply these things...it is only linear maps to and from the same space that can be composed.2012-06-12

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