Derivative of a vector Let
$p, v :$ real, positive $1\times n$ vectors,
$c^T:$ real, non - negative $n\times 1$ vector,
$I:$ the identity matrix.
Assume that the following relationship holds true:
$$p(v) = v\cdot ( I - c^Tv)^{-1}$$ 
How can we compute the derivative:
$$\dfrac{dp}{dv}(v)?$$
 A: Instead of row vectors, let's write the problem using column vectors (expressions involving row vectors just look strange to me).
Then we'll take the differential of $p$, and then rearrange it until we isolate the gradient:
$$\eqalign{
 p^T &= v^T(I-cv^T)^{-1} \cr
 p &= (I-vc^T)^{-1}\,v \cr\cr
dp &= (I-vc^T)^{-1}dv - (I-vc^T)^{-1}\,d(-vc^T)\,(I-vc^T)^{-1}v \cr
  &= (I-vc^T)^{-1}dv + (I-vc^T)^{-1}\,dv\,c^Tp \cr
  &= (1+c^Tp)\,(I-vc^T)^{-1}\,dv \cr\cr
\frac{\partial p}{\partial v} &= (1+c^Tp)\,(I-vc^T)^{-1} \cr
}$$
A: 
Disclaimer: This answer was downvoted with no explanation 18 months after being posted. Since it is correct and answers the question as formulated at the time, I guess one should consider such erratic downvotes as an inherent part of the math.SE experience. In any case... happy reading!

The gradient $\nabla p(v)$ of $p$ at $v$ such that $I-c^Tv$ is invertible is the linear function $L_v:\mathbb R^n\to\mathbb R^n$ defined, for every $w$ in $\mathbb R^n$, by $$L_v(w)=w\cdot(I-c^Tv)^{-1}+v\cdot(I-c^Tv)^{-1}\cdot c^Tw\cdot(I-c^Tv)^{-1}.$$ To prove this, try to reach a formula $$p(v+hw)=p(v)+hL_v(w)+o(\|h\|),$$ for some linear function $L_v$, when $h\to0$, $h\in\mathbb R$.
The linear function $L_v$ can be rewritten as 
$$
L_v(w)=w\cdot(I-c^Tv)^{-1}+p(v)\cdot c^Tw\cdot(I-c^Tv)^{-1}=\alpha\, w\cdot(I-c^Tv)^{-1}$$ where $\alpha$ denotes the scalar $$\alpha=1+p(v)c^T$$ hence the gradient $\nabla p(v)$ can be identified with the matrix $$M_v=\alpha\,(I-c^Tv)^{-1}=(1+p(v)c^T)\,(I-c^Tv)^{-1}$$ in the sense that, for every $w$, $$L_v(w)=w\cdot M_v$$
A: Hints:
(1) Derivative of the product of two functions:
Let $X \in \mathbb{R}^{N\times Q}$ and $F:\mathbb{R}^{N\times Q} \rightarrow \mathbb{R}^{M\times P}$ given by 
$$F(X) = G(X)\cdot H(X),$$ where $G:\mathbb{R}^{N\times Q} \rightarrow \mathbb{R}^{M\times R}$, and $H:\mathbb{R}^{N\times Q}\rightarrow \mathbb{R}^{R\times P}.$ Then,
$$\frac{dF}{dX} = \left(H^{T} \otimes I_M \right)\frac{dG}{dX} + \left(I_P \otimes G \right) \frac{dH}{dX}.$$
(2) Chain rule:
Let $X \in \mathbb{R}^{N\times Q}$ and $H:\mathbb{R}^{N\times Q} \rightarrow \mathbb{R}^{R\times S}$ such that
$$ H(X) =G(F(X)),$$ where $F:\mathbb{R}^{N\times Q} \rightarrow \mathbb{R}^{M\times P}$, and $G:\mathbb{R}^{M\times P} \rightarrow \mathbb{R}^{R\times S}.$ Then,
$$ \frac{dH}{dX} = \frac{dG}{dF}\cdot\frac{dF}{dX}.$$
(3) Useful derivatives:
(a) $$\frac{d\mathbf{x}}{d\mathbf{x}} = I, \mathbf{x} \in \mathbb{R}^{N}.$$
(b) $$ \frac{dX^{-1}}{dX} = -(X^{T})^{-1} \otimes X^{-1}, X \in \mathbb{R}^{N\times N}.$$
(c) $$ \frac{d (\mathbf{a}^{T}\mathbf{x})}{d\mathbf{x}} = \mathbf{a}^{T}, \text{for} \hspace{2mm} \mathbf{a},\hspace{0.5mm}\mathbf{x} \in \mathbb{R}^{N}.$$
(4) Useful equivalence:
$$\frac{dF}{dX} \equiv \frac{d vec(F)}{dvec(X)}.$$
