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If I have $\vec{x} = \begin{pmatrix}x_1\\ x_2\\ \vdots\\ x_n\end{pmatrix}$, how can I find the derivative of $\vec{x}^T$?

This questions comes when I was trying to find the minimum of the inner product of $\vec x$ and $A \vec x$. So I plan to differentiate $(\vec{x}, A\vec{x})$ with respect to $\vec{x}$ and find the zero. I could multiply out $(\vec{x}, A\vec{x})$ and differentiate, but I really wonder if I can do by product rule:

$$\frac{\partial}{\partial \vec x}(\vec{x}, A\vec{x}) = \frac{\partial}{\partial \vec x}(\vec{x}^TA\vec{x}) = \frac{\partial}{\partial \vec x} \vec{x}^T A\vec{x} + \vec{x}^T A \frac{\partial}{\partial \vec x} \vec{x}.$$

If I want to compute the derivative of $\vec x$ with respect to $\vec x$, I would $$\frac{\partial}{\partial \vec x} \vec{x} =\begin{pmatrix}\frac{\partial}{\partial x_1} x_1 &\frac{\partial}{\partial x_2} x_1 & \cdots & \frac{\partial}{\partial x_n} x_1\\ \frac{\partial}{\partial x_1} x_2 &\frac{\partial}{\partial x_2} x_2 & \cdots & \frac{\partial}{\partial x_n} x_2\\ \vdots & \vdots &\ddots & \vdots\\ \frac{\partial}{\partial x_1} x_n & \frac{\partial}{\partial x_2} x_n & \cdots & \frac{\partial}{\partial x_n} x_n \end{pmatrix} = \begin{pmatrix} 1 & 0 & \cdots & 0\\ 0 & 1 & \cdots & 0\\ \vdots & \vdots &\ddots & \vdots\\ 0 & 0 & \cdots & 1\\ \end{pmatrix} $$

But how can I compute $\frac{\partial}{\partial \vec x} x^T$?

Thank you

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  • 4
    $\begingroup$ $\frac{d}{dx}x$ is not even proper terminology. What exactly are you asking for, the Jacobian? The Jacobian is just the identity matrix in this case, of course. $\endgroup$ – Michael Grant Nov 18 '14 at 17:39
  • $\begingroup$ When it is ment as a symbol of differenciation in terms of how x is changes with change of x, than the result should be identity... $\endgroup$ – V-X Nov 18 '14 at 17:42
  • $\begingroup$ Thank you @MichaelGrant, I want to differentiate the inner product of $x$ and $Ax$ with respect to $x$. I could multiply out and differentiate, but I really wonder if I can do by product rule? $\endgroup$ – 1LiterTears Nov 18 '14 at 17:43
  • $\begingroup$ OP was certainly thinking about something completely different. However, the proper formalization of the question and a bit of knowledge about the problem would help... $\endgroup$ – V-X Nov 18 '14 at 17:44
  • $\begingroup$ That should have been your question in the first place, @1LiterTears. $\endgroup$ – Michael Grant Nov 18 '14 at 17:44

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