Partial Derivatives : Show that $\frac{∂x}{∂y}\frac{∂y}{∂z}\frac{∂z}{∂x}=-1$

Question

Let $f : \mathbb{R}^3 \rightarrow \mathbb{R}$. How can i show that

If $f(x,y,z)=0$ then $$\frac{∂x}{∂y}\frac{∂y}{∂z}\frac{∂z}{∂x}=-1 $$

Any help will be appreciated.

Answer 1

To understand why $$ \frac{\partial x}{\partial y}\frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = -1$$ One need to know what those symbols mean!

In above expression, $\frac{\partial x}{\partial y}$ is a short hand for $\frac{\partial X(y,z)}{\partial y}$ where $X(y,z)$ is the value of $x$ which solves the equation $f(x,y,z) = 0$ for given $y,z$. i.e., $X(y,z)$ is a function which satisfies $f(X(y,z),y,z) = 0$.

Partial differentiate against $y$ and apply chain rule, one get

$$\frac{\partial X(y,z)}{\partial y}\frac{\partial f}{\partial x}(x,y,z)|_{x=X(y,z)} + \frac{\partial f}{\partial y}(x,y,z)|_{x=X(y,z)} = 0 \quad\implies\quad \frac{\partial x}{\partial y} \stackrel{def}{=} \frac{\partial X(y,z)}{\partial y} = - \frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}$$ By a similar argument, we have

$$\frac{\partial y}{\partial z} = -\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}} \quad\text{ and }\quad \frac{\partial z}{\partial x} = - \frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial z}}$$

Multiply these $3$ relations together, the identity follows:

$$\frac{\partial x}{\partial y}\frac{\partial y}{\partial z} \frac{\partial z}{\partial x} = \left(-\frac{\frac{\partial f}{\partial y}}{\frac{\partial f}{\partial x}}\right) \left(-\frac{\frac{\partial f}{\partial z}}{\frac{\partial f}{\partial y}}\right) \left(-\frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial z}}\right) = -1$$