Logarithmic Differentiation for Multivariable functions? I was wondering if, and how exactly logarithmic differentiation may be applied to a multivariate function.
For example, I have been working with the function
$$f(x,y)=\frac{x-y}{x+y}$$
Does the following process hold?  Specifically, does the chain rule step with del f hold?
$$\ln(f(x,y))=\ln(x-y)-\ln(x+y)$$
$$\Rightarrow \frac{1}{f(x,y)} \nabla f = \langle \frac{2y}{x^2-y^2} , \frac{-2x}{x^2-y^2} \rangle$$
Solving for $\nabla f$, does give me the correct answer.  I'm wondering is my notation would be correct, and if this will ever not work.
Thanks for any help.
 A: You define
$$g(x,y)=\ln(f(x,y))=\ln(x-y)-\ln(x+y)$$
and thus
$$\nabla g=\left(\frac{1}{f(x,y)}f_x,\frac{1}{f(x,y)}f_y\right)=\frac{1}{f(x,y)}\left(f_x,f_y\right)=\frac{1}{f(x,y)}\nabla f$$
A: You could take a simpler function and checked whether it holds. The usual  differentiation rules hold true for the gradient as well:  sum, difference, product and quotient. For instance,  $\nabla(fg)=g\nabla f+f\nabla g$, $\nabla(\frac{f}{g})=\frac{g\nabla f-f\nabla g}{g^2}$, $\nabla(\ln f)=\frac{\nabla f}{f}$.
A: Yes I really believe your result as well as your notation are correct. May be you just need to put it in a more general setting. I would suggest you to proceed as follows.
Suppose that $f:\mathbb{R}^n\to\mathbb{R}$, which is differentiable and strictly positive in a certain open set $U\subset\mathbb{R}^n$. Then the function $F=\ln\circ f$ is also differntiable on $U$ with: $\nabla F(x)=\ln'(f(x))\cdot \nabla f(x)=\frac{1}{f(x)}\cdot\nabla f(x)$ (chain rule). Hence we can solve for $\nabla f(x)$, as we are in vector space $\mathbb{R}^n$ and get the answer right. 
Remember the chain rule statement:
If $G$ and $H$ are differentianble functions and $F=H\circ G$ is well defined, then $F$ is also differentiable with $\nabla F(x)=\nabla H(G(x))\cdot\nabla G(x)$. 
