What is the Process and Meaning of Taking the Derivative of a Vector?

[hkBattousai]

Here is a snapshot from one of my textbooks:
[PLAIN]http://img64.imageshack.us/img64/8114/vector0.png

How do we take the derivative below?
$\frac{d}{dx}\Huge(\normalsize x^TA^TAx\,-\,2x^TA^Tb \Huge)\normalsize\,=\,2A^TAx\,-\,2A^Tb$

There is also another vector derivative in the book as follows:
$\frac{d}{dx}\Huge(\normalsize x^Tx \Huge)\normalsize \, = \, 2x^T$

How do we take these type of derivatives?
What is the meaning of taking derivative of a vector, or transpose of vector?

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EDIT: I found http://en.wikipedia.org/wiki/Matrix_calculus#Derivative_of_linear_functions", but it doesn't either explain the main idea behind vector derivation.

$\frac{\partial \; \textbf{a}^T\textbf{x}}{\partial \; \textbf{x}} = \frac{\partial \; \textbf{x}^T\textbf{a}}{\partial \; \textbf{x}} = \textbf{a}$

$\frac{\partial \; \textbf{A}\textbf{x}}{\partial \; \textbf{x}} = \frac{\partial \; \textbf{x}^T\textbf{A}}{\partial \; \textbf{x}^T} = \textbf{A}$

 

[monea83]

For a symmetric matrix B (in your case, $$B = A^T A$$), the following is a scalar-valued function from R^n to R:

$$f(x) = x^T B x$$

The derivative you are looking for is defined as the vector of partial derivatives (aka gradient):

$$\frac{df}{dx} = \left(\frac{\partial f}{\partial x_1}, ..., \frac{\partial f}{\partial x_n}\right)^T$$

If you express f in terms of the components of B,

$$f(x) = b_{11} x_1^2 + 2 b_{12} x_1 x_2 + ...$$

you will find that the partial derivatives just "come out right", i.e.

$$\frac{df}{dx} = 2 B x$$