-----
> [!proposition] Proposition. ([[derivative of Euclidean norm squared]])
> Let $x \in \rrn$. Then $\|x\|_{2}^{2}$ is [[derivative|differentiable]] and $D(\|x\|^{2}) = 2x ^{\top}$.
>
> [!proof]- Proof. ([[derivative of Euclidean norm squared]])
>
We compute $\begin{align}D(\|x\|^{2}) = & \begin{bmatrix}
\frac{ \partial \|x\|^{2} }{ \partial x_{1} } & \dots & \frac{ \partial \|x\|^{2} }{ \partial x_{n} } \\
\end{bmatrix} \\
= & \begin{bmatrix}
\frac{ \partial (x_{1} ^{2} + \dots + x_{n}^{2} )}{ \partial x_{1} } & \dots & \frac{ \partial ( x_{1} ^{2} + \dots + x_{n}^{2} )}{ \partial x_{n} }
\end{bmatrix} \\
= & \begin{bmatrix}
2x_{1} & \dots & 2x_{n} \\
\end{bmatrix}. \\
=& 2x^{\top}.
\end{align}$
\
and doing so is justified because the third equality shows that all of $\|x\|^{2}