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I know that the derivative of $f(x)$ defined as: $\, f(x)=\dfrac{1}{2} \cdot \left|\left|g(x) \right|\right|^2_2$

is $\nabla f(x) = J_g(X)^T \cdot g(x)$

where $g:\mathbb{R}^n \rightarrow \mathbb{R}^m$, $f:\mathbb{R}^n \rightarrow \mathbb{R}$, and $\left|\left|x \right|\right|_p$ is the $p$-norm of $x$ : $\left|\left|x \right|\right|_p=\left( \sum\limits_{i=1}^n \left|x_i \right|^p \right)^\frac{1}{p}$

First of all why is the norm sign removed after derivation?

Secondly, what is the derivative in for example this case where the norm and exponent differ:

$f(x)=\dfrac{1}{2} \cdot \left|\left|g(x) \right|\right|^8_3$

Could you please explain the reason behind the two cases?

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1) The norm sign is removed after derivation because you would expect $\nabla f$ to be a vector, and $\Vert g(x)\Vert_2$ is a scalar.

You might think about it this way: Let $n\colon \mathbb{R}^m$ be the (square of the) norm function: $$ n(y) = \Vert y \Vert_2^2 = \sum_{i=1}^m y_i^2 $$ Then for each $j$, $$ \frac{\partial n}{\partial y_j} = 2 y_j $$ which tells you that $\nabla n(y) = 2y$.

Your function is $f(x) = \frac{1}{2} n(g(x))$, so by the Chain Rule, $$ \nabla f(x) = \frac{1}{2} \nabla n(g(x)) \cdot J_g(x) = \frac{1}{2} \cdot 2 g(x) \cdot J_g(x) = g(x) \cdot J_g(x) $$ modulo transpositions that's what you want.

2) This is harder than I originally thought, partly because the $p$-norm isn't a differentiable function for all $y$. But whatever you can work out is still due to the Chain Rule.

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