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Wikipedia lists the identity for the gradient of a composition as $$\nabla(f\circ \mathbf A) = (\nabla f\circ \mathbf A)\nabla \mathbf A$$

First, is this formula correct?

Assuming it is:

Second, what does $\nabla\mathbf A$ mean? Is it the Jacobian of $\mathbf A$?

And finally, what does it mean to multiply the vector $\nabla f\circ \mathbf A$ and the matrix $\nabla \mathbf A$? Is it some type of tensor product? That doesn't seem like it would result in the correct type of object. Or is $\nabla f\circ \mathbf A$ supposed to be a row matrix and then you just use matrix multiplication?

Bobbie D
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    The formula is correct with $\nabla A$ the Jacobian matrix. It's the usual product of matrices. $\nabla f(A(x))$ is a $1\times n$ matrix, while $\nabla A(x)$ is a a $n\times s$ matrix, resulting in a $1\times s$ matrix. –  Oct 16 '16 at 20:56
  • OK, cool. Thanks. – Bobbie D Oct 16 '16 at 20:57
  • This is simply the chain rule stated in terms of $\nabla$ operator. – amd Oct 16 '16 at 21:05

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