How is it that this partial derivative is equal to 0?

Question

In my probability textbook are the following equations:
$Z=X/Y$ and $X=W$.

There is a Jacobian as follows:

$$J = \begin{bmatrix} \frac{\partial x}{\partial z} & \frac{\partial x}{\partial w} \\[1ex] % <-- 1ex more space between rows of matrix \frac{\partial y}{\partial z} & \frac{\partial y}{\partial w} \\[1ex] \end{bmatrix}$$

solving for the partial derivatives (according to my textbook):

$$J = \begin{bmatrix} 0 & 1 \\[1ex] % <-- 1ex more space between rows of matrix -\frac{w}{z^2} & \frac{1}{z} \\[1ex] \end{bmatrix}$$

How is it that $\frac{\partial x}{\partial z} = 0?$

My thinking is $X=YZ$ so $\frac{\partial x}{\partial z} = Y$. Where am I going wrong? All the other partial derivatives make perfect sense to me.

Thanks in advance.

Answer 1

Think of $x$ as a function of two variables $z$ and $w$. Then $x(z,w)=w$. Since $z$ doesn't appear in this expression, you have $\partial x/\partial z = 0$.

The point is to express $x$ and $y$ as functions of $z$ and $w$. Then regarding $x$ and $y$ as the dependent variables, and $z$ and $w$ as the independent variables, compute the partials $\tfrac{\partial}{\partial z}$ and $\tfrac{\partial}{\partial w}$ of $x=x(z,w)$ and of $y=y(z,w)$.

Answer 2

The Jacobean concerns $x,y$ as functions of $z,w.$ So the equations to work with are $x=w$ and $y=\frac{w}z.$