Derivative of $x \mapsto \operatorname{tr}((A-xx^T)^{1/2})$ Let $A$ be an $n \times n$ positive definite matrix and let $x \in \mathbb{R}^n$ such that $A-xx^T$ is positive semidefinite. 
I would like to find the derivative of
$$x \mapsto \operatorname{tr} \left( (A-xx^T)^{1/2} \right)$$
 A: There is an important point.
Proposition. Let $f$ be an analytic function and let $\phi:U\in Z\subset M_n\rightarrow trace(f(U))$. Then $D\phi_U:H\in M_n\rightarrow trace(f'(U)H)$.
Proof. Locally, $f(U)=\sum_{i=0}^\infty u_iU^i$ and 
$D\phi_U(H)=\sum_{i=0}^\infty u_itrace(D(U^i)(H))=\sum_{i=0}^\infty u_i\;i\;trace(U^{i-1}H)=trace(f'(U)H)$.  $\square$
Application. Let $\phi: U\in S_{>0}\rightarrow trace(U^{1/2})$ where $S_{>0}$ is the set of $>0$ symm. matrices. Then $D\phi_U(H)=1/2trace(U^{-1/2}H)$.
Here we consider $g=\phi\circ f:x\rightarrow trace((A-xx^T)^{1/2})$ where $f:x\rightarrow A-xx^T$ and $A-xx^T>0$.
Then $Dg_x:h\in\mathbb{R}^n \rightarrow 1/2trace((A-xx^T)^{-1/2}(-hx^T-xh^T)  )=-trace(x^T(A-xx^T)^{-1/2}h)$. 
If you prefer the gradient, then we deduce $\nabla g(x)=-(A-xx^T)^{-1/2}x$.
Remark. Of course, I find the same result as @greg.
A: Assume the matrix $$M=A-xx^T$$ is positive definite, and let
$$\eqalign{
  B &= \frac{1}{2}M^{-1/2}
}$$
Then  you can find the differential and gradient of the function as
$$\eqalign{
 \phi &= {\rm tr}(M^{1/2}) \cr
d\phi &= B^T:dM = -B^T:(dx\,x^T+x\,dx^T) = -(B+B^T)x:dx \cr
\frac{\partial\phi}{\partial x} &= -(B+B^T)x \cr
}$$
where colon was used as a handy product notation for the trace, i.e. 
$$A:B={\rm tr}(A^TB)$$
