Derivative with Respect to a Ratio of Variables We have two strictly positive real-valued variables $x$ and $y$ and a third one defined as $z=\frac{x}{y}$.
Question 1: How do I compute the derivative $\frac{\partial x}{\partial z}$? Is it $\frac{\partial x}{\partial z} = \frac{1}{\frac{\partial z}{\partial x}}=\frac{1}{\frac{1}{y}}=y?$
Question 2: What is the derivative $\frac{\partial (xz)}{\partial z}$? Do I use the answer to Question 1 and the product rule to get $\frac{\partial x}{\partial z}z+x\frac{\partial z}{\partial z}=yz+x=y\frac{x}{y}+x=2x$? Or is it something else?
Thanks!
 A: *

*Let $f(x,y,z)=z-\frac{x}{y}$. Fix $y_0\not=0,z_0\in\mathbb{R}$ and consider the equation $f(x,y_0,z_0)=0$. Using the implicit function theorem (http://en.wikipedia.org/wiki/Implicit_function_theorem):
$$(Jf)(x_0,y_0,z_0)=\left(\matrix{-\frac{1}{y} & \frac{1}{y^2} & 1} \right)|_{y=y_0}=\left(\matrix{-\frac{1}{y_0} & \frac{1}{y_0^2} & 1} \right)$$
so $\exists g:\mathbb{R}^2\to\mathbb{R}$ such that $f\big(g(y,z),y,z\big)=0$ for $y,z$ in a neighbourhood of $y_0,z_0$ (it's clear that $g(y,z)=zy)$. But the theorem also states that $Jg=-\left(\frac{1}{-1/y_0}\right)(Jf)=y_0\cdot Jf$ so
$$\frac{\partial g}{\partial z}=y_0\cdot\frac{\partial f}{\partial z}=y_0$$

*For the second point we are still working in a neighbourhood of $(y_0,z_0)$:
$$\frac{\partial}{\partial z}zg=z\frac{\partial}{\partial z}g+g\frac{\partial}{\partial z}z=zy_0+g$$
so evaluating in $(y_0,z_0)$ 
$$\frac{\partial(zg)}{\partial z}(y_0,z_0)=z_0 y_0+g(y_0,z_0)=z_0 y_0+z_0 y_0=2g(y_0,z_0)$$
which is what you proposed in your question.
