Solving Gradient of ||f(x)||^2 - Chain Rule

[perplexabot]

Hey,
I have been trying to figure out how to solve $\triangledown_x ||f(x)||^2_2$.
I have used the chain rule (hopefully correctly) to get the following:
$$\triangledown_x ||f(x)||^2_2=2\triangledown_xf(x)^T \frac{f(x)^T}{||f(x)||_2}$$
Is this correct?

The reason I doubt my answer is because I know the gradient of a scalar valued function should be a vector. My answer seems to give a scalar. Can anyone please shed some light...

Note: $x \in \Re^n$ and I am using the convection that the gradient, $\triangledown_x$, of a function is a row vector. Also assume $f: \Re^n\rightarrow \Re^m$ .

 

[perplexabot]

So, I was washing the dishes when I realized that $$2\triangledown_xf(x)^T \frac{f(x)^T}{||f(x)||_2}$$ is not a scalar. This is because $\triangledown_xf(x)^T$ is a matrix (this is actually the jacobian!). So now I have a feeling the above may be close, but still wrong. I would appreciate a confirmation tho.

Thank you :)

EDIT: Just realized this is my 300th post! YAY me! Love you physics forums!

 

[perplexabot]

So I have made a couple of fixes, specifically making sure the matrix multiplaction agrees "dimensionally" $$\triangledown_x ||f(x)||^2_2=\frac{2f(x)^T}{||f(x)||_2}(\triangledown_xf(x))$$
I wonder if this is correct now. Anyone?

 

[perplexabot]

My work is wrong! Here is the correct method for the sake of completeness.
$$\begin{equation*}<br /> \begin{split}<br /> \triangledown_x ||f(x)||^2_2 &=\triangledown_x (f(x)^Tf(x)) \\<br /> &=(\triangledown_xf(x)^T)^Tf(x)+(\triangledown_xf(x))^Tf(x) \\<br /> &=(\triangledown_xf(x))^Tf(x)+(\triangledown_xf(x))^Tf(x) \\<br /> &=2(\triangledown_xf(x))^Tf(x)<br /> \end{split}<br /> \end{equation*}$$

Thank you all for reading.
EDIT: Also here and here are some references I used.