How to calculate derivative of a vector valued function

[Johnson04]

Homework Statement

\vec f_1 : \bf R^n \rightarrow R^1, \vec f_1(\vec x) = \|\vec x\|^2, calculate \vec f_1'(\left[ \begin{array}{c} 1 \\\vdots\\ 1 \end{array} \right])

Homework Equations

N/A

The Attempt at a Solution

I calculated the derivative as follow:
Since
\vec f_1(\vec x) =\|\vec x \|^2, \vec f_1(\vec x) = \sum \limits_{i=1}^n x_i^2.
So
\vec f_1(\vec x + \vec h) = \sum \limits_{i = 1}^n (x_i + h_i)^2 = \sum \limits_{i=1}^n (x_i^2 + 2x_i h_i + h_i^2) = \sum \limits_{i=1}^n x_i^2 + \sum \limits_{i=1}^n 2x_i h_i + \sum \limits_{i=1}^n h_i^2.
Thus, we have:
\vec f_1(\vec x + \vec h) - \vec f_1(\vec x) = \sum \limits_{i=1}^n (2x_i h_i +h_i^2)

Since according to the definition of the derivative of vector valued function (Rudin Analysis Page 212):
If there exists a linear transformation A such that
\lim \limits_{\vec h to 0} \frac{|\vec f(\vec x + \vec h) - \vec f(\vec x) - A\vec h |}{|\vec h|} = 0,
then \vec f is differentiable at \vec x, and
\vec f'(\vec x) = A

I defined a linear transformation as: A = 2 \vec x^T, such that we may have A \vec h = \sum \limits_{i = 1}^n 2x_i h_i. Then since \lim \limits_{\vec h to 0} \frac{|\sum \limits_{i = 1}^n h_i^2|}{|\vec h|} = \lim \limits_{\vec h to 0} |\vec h| = 0, A = 2 \vec x^T = \vec f'(\vec x), and then plug \vec x = \begin{pmatrix} 1 \\ \vdots \\1 \end{pmatrix} into the expression \vec f'(\vec x).

Is this solution correct? Thanks!

 

[HallsofIvy]

A function from Rⁿ to R is essentially a function f such that f(x_1, x_2, x_3,\cdot\cdot\cdot, f_n) is a number for each (x_1, x_2, \cdot\cdot\cdot, x_n) Its derivative can represented by the vector
<\frac{\partial f}{\partial x_1}, \frac{\partial f}{\partial x_2}, \cdot\cdot\cdot, \frac{\partial f}{\partial x_n}>
the gradient of the function.

||x||^2= x_1^2+ x_2^2+ \cdot\cdot\cdot+ x_n^2
The gradient is 2x_1+ 2x_2+ \cdot\cdot\cdot+ 2x_n

Evaluate that at x_1= x_2= x_3= \cdot\cdot\cdot= x_n= 1.

To answer your specific question,
\lim \limits_{\vec h to 0} \frac{|\sum \limits_{i = 1}^n h_i^2|}{|\vec h|}
is 0 but that omly tells you the derivative at (0, 0, ..., 0), not at (1, 1, 1,...,1).
What is
\lim \limits_{\vec h to 0} \frac{|\sum \limits_{i = 1}^n (1+h_i)^2- 1|}{|\vec h|}?

 

[Johnson04]

Thanks for replying!
\lim \limits_{\vec h \rightarrow 0} \frac{|\sum \limits_{i=1}^n h_i^2|}{|\vec h|} was obtained from \lim \limits_{\vec h \rightarrow 0} \frac{|\vec f(\vec x + \vec h) - \vec f(\vec x) - A\vec h|}{|\vec h|}. Since I defined the linear transform at \vec x = \left[\begin{array}{c} x_1 \\ \vdots \\ x_n \end{array}\right] as A = \left(\begin{array}{ccc} 2x_1 & \ldots & 2x_n \end{array}\right), I got A\vec h = \sum \limits_{i = 1}^n 2x_i h_i.

Plug this A\vec h back to the definition of derivative, I got:
\lim \limits_{\vec h \rightarrow 0} \frac{|\vec f(\vec x + \vec h) - \vec f(\vec x) - A\vec h|}{|\vec h|} = \lim \limits_{\vec h \rightarrow 0} |\vec h| = 0..

I think I have shown that the linear transformation A does exist at \vec x, so according to the definition \vec f'(\left[\begin{array}{c} 1 \\ \vdots \\ 1 \end{array}\right]) = \left(\begin{array}{ccc} 2 & \ldots & 2 \end{array}\right).

But I just was confused by the fact that the function \vec f(\vec x) actually maps \bf R^n to \bf R^1, but the result of my calculation implies that \vec f'(\vec x) maps \bf R^n to itself? Am I wrong somewhere? Or I still didn't really understand the definition?
Thanks!