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I've got a function $g(x,y) = || f(x,y) ||_2 $ and I want to calculate its derivatives with respect to $x$ and $y$.

Using Mathematica, differentiating wrt $x$ gives me $ f'_x(x,y) \text{Norm}'( f(x,y))$, where Norm is $|| \ldots ||$.

I read here that $d||{\bf x}|| = \frac{ {\bf x}^Td{\bf x}}{||{\bf x}||}$ (at least for 2-norm)

Point is, as inside the norm I have a multivariate function, I'm still confused on how to calculate $ f'_x(x,y) \text{Norm}'( f(x,y))$

I think it should be $f'_x(x,y) \frac{f(x,y)}{||f(x,y)||}$, but some verification would be great :)

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  • $\begingroup$ When you are not sure, writing out all components should help. It's nasty, I know, but it always works. $\endgroup$ – Tunococ Jan 31 '13 at 12:30
  • $\begingroup$ First find the derivative of $h(x,y) = \|f(x,y)\|^2 = f(x,y) \cdot f(x,y)$, using the product rule. Then you can get the derivative of $g(x,y) = \sqrt{h(x,y)}$ from the chain rule. $\endgroup$ – Hans Engler Jan 31 '13 at 12:36
  • $\begingroup$ Just saw your edit. This answer is correct. $\endgroup$ – Hans Engler Jan 31 '13 at 12:36
  • $\begingroup$ @YellowSkies: how is your link related to this question? $\endgroup$ – Watson Sep 14 '16 at 9:14
  • $\begingroup$ @Tunococ can you solve this one ? math.stackexchange.com/questions/2587031/… thanx. $\endgroup$ – Usman Mahmood Dec 31 '17 at 21:23
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Suppose $f:\mathbb R^m \to \mathbb R^n$. Decompose into $f = (f_1, \ldots, f_n)$. Each $f_i$ is a real-valued function, i.e., $f_i: \mathbb R^m \to \mathbb R$. Then $$ g(X) = \|f(X)\|_2 = \sqrt{\sum_{i=1}^n f_i(X)^2}. $$ Therefore, $$\nabla g(X) = \frac 12\left(\sum_{i=1}^n f_i(X)^2\right)^{-\frac 12}\left(\sum_{i=1}^n 2f_i(X)\nabla f_i(X)\right) = \frac{\sum_{i=1}^n f_i(X)\nabla f_i(X)}{\|f(X)\|_2}. $$ This matches your answer.

If you want to write in terms of the Jacobian matrix of $f$ instead of components $f_i$, you can: $$ \nabla g(X) = \frac{J_f(X)^T f(X)}{\|f(X)\|_2}. $$

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  • $\begingroup$ Many thanks for the answer! $\endgroup$ – Babis Jan 31 '13 at 13:14

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