Derivative of the nuclear norm The nuclear norm is defined in the following way
$$\|X\|_*=\mathrm{tr} \left(\sqrt{X^T X} \right)$$
I'm trying to take the derivative of the nuclear norm with respect to its argument
$$\frac{\partial \|X\|_*}{\partial X}$$
Note that $\|X\|_*$ is a norm and is convex. I'm using this for some coordinate descent optimization algorithm. Thank you for your help.

Alt's answer has a fundamental error. First of all, the nuclear norm is the sum of all singular values not the absolute of the singular values. 
To make it right, we need to first define the square root for matrix as $\sqrt{BB}=B$. As Alt shown,
$||x||_*=tr(\sqrt{V\Sigma^2V^T})$
But we cannot use the circularity of trace here because it is not well defined.  
We should do something like this, 
$||x||_*=tr(\sqrt{V\Sigma^2V^T})=tr(\sqrt{V\Sigma V^TV\Sigma V^T})=tr(V\Sigma V^T)$, 
the last equality is based on the definition of the square root for matrix described above. Then by the circularity of trace, we get
$tr(V\Sigma V^T)=tr(\Sigma V^TV)=tr(\Sigma)=\sum_i \sigma_i$.
