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 w.r.t. $x$ gives me $ f'_x(x,y) \text{Norm}'( f(x,y))$, where Norm is $\| \cdot \|$.

I read here that

$$d\|{\bf x}\| = \frac{ {\bf x}^Td{\bf x}}{\|{\bf x}\|}$$

at least for the $2$-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 :)

Solution 1:

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