Differentiation of functions w.r.t. a composed argument I need help with the following derivative involving inner products: $$\frac{d\, \log(x)^T\,y}{d\,x^T\,y}$$
Here $x$ and $y$ are $n$-dimensional vectors, $T$ indicates transpose, and the logarithm of the vector is taken point-wise: $\log(x)_i = \log(x_i)$.
THANKS for your help in advance!!!
 A: $(\log x)^T*y=\log(x_1^{y^1}.x_2^{y_2}.\ldots.x_n^{y_n})$ So you have to differentiate 
$x_1^{y_1}.x_2^{y_2}.\ldots.x_n^{y_n}$ with respect to $x_1y_1+x_2y_2+\ldots+x_ny_n$.
A: $\def\nR{\mathbb{R}}\def\sign{\operatorname{sign}}\def\l{\left}\def\r{\right}\def\ltag#1{\tag{#1}\label{#1}}$
The only interpretation of the question I can think of is that you are given a vector of unknowns $z=[x,y]$ and two functions $f(z)=\log(x)^T\cdot y$ and $g(z)=x^T y$ and you should determine $a(z)$ such that
$$
\ltag{quest}
\frac{df(z)}{dg(z)} = a(z)
$$
where $df$ and $dg$ are to be interpreted as differentials. In this case you have
\begin{align}
df(z) &= a(z)\cdot dg(z)\\
f'(z)\cdot dz &= a(z)\cdot g'(z)\cdot dz\\
(f'(z)-a(z)g'(z))\cdot dz = 0
\end{align}
with $f'(z),g'(z)\in \nR^{1\times N}$. Note, that this equation is a generalized eigenvalue problem with $a(z)$ as the eigenvalue and $dz\in\nR^N$ as eigen direction.
In our case it is just a single equation which we can tackle directly.
\begin{align}
f'(z) &= \begin{pmatrix}
(y./x)^T,&\log(x)^T
\end{pmatrix}\\
g'(z) &= \begin{pmatrix}
y^T&x^T
\end{pmatrix}
\end{align}
We have to satisfy the equation
\begin{align}
\ltag{d}
((y./x)^T-a(z)\cdot y^T)\cdot dx + (\log(x)^T-a(z)\cdot x^T)\cdot dy &= 0
\end{align}
Let us assume the differentials $d x,dy$ were not restricted. Then the equations
\begin{align}
((y./x)^T-a(z)\cdot y^T)&=0\ltag{allDir1}\\
(\log(x)^T-a(z)\cdot x^T)&=0\ltag{allDir2}
\end{align}
would all have to be satisfied. The first component of equation \eqref{allDir1} reads as
$$
\ltag{a1}
a = \frac1{x_1}
$$
But, the other components of \eqref{allDir1} imply $x_k=\frac1{a}=x_1$ for $k=2,\ldots,n$ and the other equation \eqref{allDir2} says
\begin{align}
\ltag{a2}
a&=\frac{\log(x_k)}{x_k}&&\text{ for }k=1,\ldots,n.
\end{align}
Equations \eqref{a1} and \eqref{a2} are not solvable together.
Therfore, all solutions of equation \eqref{d} restrict the differential $dz$.
I think equation \eqref{quest} imposes further $g'(z)dz\neq 0$.
One could ask which differentials $dz$ and which values $a$ are admissible in this case.
For an instance the hyperplane $y=0$ is not admissible even if it looks good for \eqref{d}. It is not admissible since with $y=0$ we also have $dy=0$ and $g'(z)dz = y^Tdx + x^Tdy =0$.
Let us look at a hyperplane with $y=c$ for some fixed $c\neq 0\in\nR^n$.
Because of $dy=0$ we need $0\neq g'(z)dz=c^Tdx$.
In this case \eqref{d} gives the equation
\begin{align}
\sum_{k=1}^n \frac{c_k dx_k}{x_k}-ac_kdx_k = 0
\end{align}
Let us try $a=0$ and let us assume that just two components of $c_k$ are nonzero, e.g., $c_1=1$ and $c_2=-1$. We would end up with the equation
\begin{align}
\frac{dx_1}{x_1} = \frac{dx_2}{x_2}.
\end{align}
With the requirement $c_1\cdot dx_1 + c_2 \cdot dx_2\neq 0$ there would locally be at least one of the differentials $dx_1 , dx_2$ nonzero (the considered tangent vector to the manifold at hand would have a nonzero component in this direction). Integration yields
\begin{align}
\int_{x_{10}}^{x_1} \frac{d \bar x_1}{\bar x_1} &= \int_{x_{20}}^{x_2} \frac{d \bar x_2}{\bar x_2}\\
\log\l(\frac{x_1}{x_{10}}\r) &= \log\l(\frac{x_2}{x_{20}}\r)\\
\ltag{sur}
\frac{x_1}{x_{10}}  &= \frac{x_2}{x_{20}}
\end{align}
One case for an admissible solution is $a=0$ with a tangent vector $dz$ on plane hyper-surfaces with $y=\begin{pmatrix}1&-1&0&\ldots&0\end{pmatrix}^T$ satisfying equation \eqref{sur} and $x_1,x_2\neq 0$.
This is not a complete characterization of the solution set of \eqref{quest}. But 1st it gives a reasonable interpretation of the question and 2nd gives some insight in the structure of the solution set of \eqref{quest}.

Note, that differentials can be interpreted as components of tangent vectors as it is described in my answer at https://physics.stackexchange.com/questions/92925/how-to-treat-differentials-and-infinitesimals/93025#93025.
A: This question isn't undefined or ambiguous at all, it's just a lot of work.  Rewrite the given derivative into an equivalent expression with no derivatives except for the $n^2$ atomic expressions $\frac{d\,x_i}{d\,y_j}$ (see footnote)
$$\frac{d\,\sum_{i=1}^n \log(x_i) y_i}{d\,\sum_{j=1}^n x_j y_j}$$
$$\frac{\sum_{i=1}^n d\,\log(x_i) y_i}{\sum_{j=1}^n d\,x_j y_j}$$
$$\sum_{i=1}^n \frac{d\,\log(x_i) y_i}{\sum_{j=1}^n d\,x_j y_j}$$
$$\sum_{i=1}^n \left(\frac{\sum_{j=1}^n d\,x_j y_j}{d\,\log(x_i) y_i}\right)^{-1}$$
$$\sum_{i=1}^n \left(\sum_{j=1}^n \frac{ d\,x_j y_j}{d\,\log(x_i) y_i}\right)^{-1}$$
$$\sum_{i=1}^n \left(\sum_{j=1}^n \frac{ x_j\,d\,y_j + y_j\,d\,x_j}{\log(x_i)\,d\, y_i + y_i/x_i\,d\, x_i}\right)^{-1}$$
$$\sum_{i=1}^n \left(\sum_{j=1}^n 
\frac{ x_j\,d\,y_j}{\log(x_i)\,d\, y_i + y_i/x_i\,d\, x_i}
+
\frac{y_j\,d\,x_j}{\log(x_i)\,d\, y_i + y_i/x_i\,d\, x_i}\right)^{-1}$$
$$\sum_{i=1}^n \left(\sum_{j=1}^n 
\left(\frac{\log(x_i)\,d\, y_i + y_i/x_i\,d\, x_i}{ x_j\,d\,y_j}\right)^{-1}
+
\left(\frac{\log(x_i)\,d\, y_i + y_i/x_i\,d\, x_i}{y_j\,d\,x_j}\right)^{-1}
\right)^{-1}$$
$$\sum_{i=1}^n \left(\sum_{j=1}^n 
\left(
\frac{\log(x_i)}{ x_j}\frac{d\, y_i}{d\,y_j}
  +
\frac{y_i/x_i}{ x_j}\frac{d\, x_i}{d\,y_j}
\right)^{-1}
+
\left(
\frac{\log(x_i)}{y_j}\frac{d\, y_i}{d\,x_j}
  +
\frac{y_i/x_i}{y_j}\frac{d\, x_i}{d\,x_j}
\right)^{-1}
\right)^{-1}$$
A special case of this (probably the one you are interested in) is where all of the variables are independent, that is:
$$(\forall i, j)\frac{d\,x_i}{d\,y_j} = 0$$
$$(\forall i \ne j)\frac{d\,x_i}{d\,x_j} = \frac{d\,y_i}{d\,y_j} = 0$$
$$(\forall i = j)\frac{d\,x_i}{d\,x_j} = \frac{d\,y_i}{d\,y_j} = 1$$
Edit: I believe the above derivation is only correct when the variables have a functional relationship, like $y = f(x)$.  When they are independent, it no longer seems to be correct, and I apologize, I haven't figured out how to solve the problem in that case yet.

Footnote:
Another way of thinking about the expression:  At any point in the logic, you should be able to apply an arbitrary set of consistent relationships between the variables and get a consistent result.  For example, consider if the poster had asked for the simpler expression:
$$\begin{align}
\frac{d\,xy}{d\,x + y} &= \frac{xd\,y + yd\,x}{d\,x + d\,y}\\
&= \frac{x\,d\,y}{d\,x + d\,y} + \frac{y\,d\,x}{d\,x + d\,y}\\
&= \left(\frac{d\,x + d\,y}{x\,d\,y}\right)^{-1} + \left(\frac{d\,x + d\,y}{y\,d\,x}\right)^{-1}\\
&= \left(\frac{d\,x}{x\,d\,y} + \frac{d\,y}{x\,d\,y}\right)^{-1}
 + \left(\frac{d\,x}{y\,d\,x} + \frac{d\,y}{y\,d\,x}\right)^{-1}\\
&= x\left(\frac{d\,x}{d\,y} + 1\right)^{-1}
 + y\left(1 + \frac{d\,y}{d\,x}\right)^{-1}\\
\end{align}$$
Continuing the example, you are still free to add a new assumption, such as $y = e^{kx}$.
$$\begin{align}
\frac{d\,xy}{d\,x + y} & = \frac{d\,xy}{d\,x} \left(\frac{d\,x + y}{d\,x}\right)^{-1}\\
&= \frac{d\,x\,e^{kx}}{d\,x} \left(\frac{d\,x + e^{kx}}{d\,x}\right)^{-1}\\
&=\frac{(k\,x\,+1)e^{kx}}{ke^{kx} + 1}
\end{align}$$
And applying the new assumption to the result:
$$\begin{align}
x\left(\frac{d\,x}{d\,y} + 1\right)^{-1}
 + y\left(1 + \frac{d\,y}{d\,x}\right)^{-1}
 & = x\left(k^{-1}e^{-kx} + 1\right)^{-1}
   + e^{kx}\left(1 + ke^{kx}\right)^{-1}\\
 & = x\left(\frac{ke^{kx}}{ke^{kx} + 1}\right) + e^{kx}\left(\frac{k^{-1}e^{-kx}}{k^{-1}e^{-kx} + 1}\right)\\
 & = kx\left(\frac{e^{kx}}{ke^{kx} + 1}\right) + e^{kx}\left(\frac{1}{1 + ke^{kx}}\right)\\
 & = \frac{(kx+1)e^{kx}}{ke^{kx} + 1}
\end{align}$$
Although this is only 1 example, it suggests that this sort of calculus is well defined and consistent.  It's just not the limited sort of calculus-of-functions that you see in schools.
