Multivariable Chain Rule Notation

Question

(without using tree diagram)

If

z = f(x − y)

g(x, y) = x − y

so that z =$f∘g$

use the chain rule to show that

$\frac{∂z}{∂x}$+ $\frac{∂z}{∂y}$ = 0

Answer: Why does the chain rule imply this answer?

$\frac{∂z}{∂x}$ =$\frac{∂f}{∂g}$$\frac{∂g}{∂x}$ =$\frac{∂f}{∂g}$

and

$\frac{∂z}{∂y}$=$\frac{∂f}{∂g}$$\frac{∂g}{∂y}$ =$-\frac{∂f}{∂g}$

Answer 1

There is no such thing as ${\partial f\over\partial x}$ or ${\partial f\over\partial y}$. We are given that $$z(x,y):=f\bigl(g(x,y)\bigr)\ ,$$ where $f$ is a function of one variable, hence has an ordinary derivative $f'$, and $g(x,y)=x-y$. The chain rule says that $${\partial z\over\partial x}=f'\bigl(g(x,y)\bigr){\partial g\over\partial x}=f'\bigl(g(x,y)\bigr)\cdot1\ ,$$ and similarly $${\partial z\over\partial y}=f'\bigl(g(x,y)\bigr){\partial g\over\partial y}=f'\bigl(g(x,y)\bigr)\cdot(-1)\ .$$ It follows that $${\partial z\over\partial x}+{\partial z\over\partial y}\equiv0\ .$$

Answer 2

Using the chain rule you have proved that: $$ \frac{\partial z}{\partial x} =\frac{\partial f}{\partial g} $$ and $$ \frac{\partial z}{\partial y} =-\frac{\partial f}{\partial g} $$

so : $$ \frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}=\frac{\partial f}{\partial g}-\frac{\partial f}{\partial g}=0 $$

And this result is true in general (if the request derivatives exist) thanks to the chain rule.