Derivative of the 2-norm of a multivariate function

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 :)

• When you are not sure, writing out all components should help. It's nasty, I know, but it always works. – Tunococ Jan 31 '13 at 12:30
• 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. – Hans Engler Jan 31 '13 at 12:36
• Just saw your edit. This answer is correct. – Hans Engler Jan 31 '13 at 12:36
• @YellowSkies: how is your link related to this question? – Watson Sep 14 '16 at 9:14
• @Tunococ can you solve this one ? math.stackexchange.com/questions/2587031/… thanx. – Usman Mahmood Dec 31 '17 at 21:23

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: $$\nabla g(X) = \frac{J_f(X)^T f(X)}{\|f(X)\|_2}.$$