Graduate Is the Chain Rule Applicable to the Euclidean Norm in Calculating Derivatives?

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The discussion centers on the application of the chain rule to the Euclidean norm in derivative calculations involving 3D vectors and a matrix. Two derivative expressions are proposed for evaluation, concerning the vector b and the matrix W, with a focus on their correctness. The operator ∂/∂b is clarified as equivalent to the gradient operator ∇b. Initial skepticism about the expressions' meaningfulness is expressed, but the conversation remains open to further exploration. The thread ultimately seeks validation of the derivative calculations in the context of vector calculus.
SchroedingersLion
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Need some verification or corrections.
Greetings,

suppose we have 3d vectors ##\mathbf{x}_k, \mathbf{y}_k, \mathbf{b}## for ##k=1,...,N## and a 3x3 matrix ##\mathbf{W}## with real elements ##w_{i,j}##.

Are the following two results correct?
$$
\frac{\partial}{\partial \mathbf{b}} \sum_k ||\mathbf{Wx}_k+\mathbf{b}-\mathbf{y}_k||² = \sum_k 2(\mathbf{Wx}_k+b-\mathbf{y}_k)
$$
$$
\frac{\partial}{\partial w_{i,j}} \sum_k ||\mathbf{Wx}_k+\mathbf{b}-\mathbf{y}_k||² = \sum_k 2 (\mathbf{Wx}_k+\mathbf{b}-\mathbf{y}_k)\cdot
\begin{pmatrix}
0 \\
... \\
0 \\
x_{k,j}\\
0\\
...\\
0
\end{pmatrix}
$$

where the nonzero entry in the column vector is in row ##i## and where ##x_{k,j}## is the ##j-th## component of vector ##\mathbf{x}_k##.
Calculating the scalar product gives
$$
\sum_k 2(\sum_{n=1}^{3} w_{i,n}x_{k,n} +b_{i} - y_{k,i})x_{k,j}
$$
 
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SchroedingersLion said:
$$
\frac{\partial}{\partial \mathbf{b}} \sum_k ||\mathbf{Wx}_k+\mathbf{b}-\mathbf{y}_k||² = \sum_k 2(\mathbf{Wx}_k+b-\mathbf{y}_k)
$$
I have never seen the operator $$ \frac{\partial}{\partial \mathbf{b}} $$ before. Can you define it?
At first glance, the expression is not meaningful - but I am willing to suspend disbelief ...
 
Svein said:
I have never seen the operator $$ \frac{\partial}{\partial \mathbf{b}} $$ before. Can you define it?
At first glance, the expression is not meaningful - but I am willing to suspend disbelief ...
Oh, that is just a way of writing ##\nabla_{\mathbf{b}}##, sorry.
 

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