Prove that $(\frac{\partial z}{\partial x})^{2}-(\frac{\partial z}{\partial y})^{2}=\frac{\partial z}{\partial a}\frac{\partial z}{\partial b}$

Question

Given $x=a+b,y=a-b$, prove that $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{\partial z}{\partial a}\frac{\partial z}{\partial b}$$

$z=f(x,y)$

It is the first time that I've seen this weird greek sign. is it something like $dx/dy$ ? what is going on in this question? I know that is $d/dy$ is more like an operator than a fraction. Or perhaps I'm just mixing up 2 different things..? sorry it just that I'm not familiar with this particular sign...

Answer 1

Hint.

$$ \frac{\partial x}{\partial a}+\frac{\partial y}{\partial a} = 2\\ \frac{\partial x}{\partial b}-\frac{\partial y}{\partial b} = 2\\ \frac{\partial x}{\partial a}-\frac{\partial y}{\partial a} = 0\\ \frac{\partial x}{\partial b}+\frac{\partial y}{\partial b} = 0\\ $$

and

$$ \frac{\partial z}{\partial x} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial x}\\ \frac{\partial z}{\partial y} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial y}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial y} $$

then

$$ \left(\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\right)\left(\frac{\partial z}{\partial x}-\frac{\partial z}{\partial y}\right)=\cdots $$

Answer 2

$x=a+b,\quad y=a-b\implies a=\frac{1}{2}(x+y)\quad \text{and}\quad b=\frac{1}{2}(x-y)$

By chain rule $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial x}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial x}$$ and $$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial a}\frac{\partial a}{\partial y}+\frac{\partial z}{\partial b}\frac{\partial b}{\partial y}$$

So $$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial a}\frac{1}{2}+\frac{\partial z}{\partial b}\frac{1}{2}=\frac{1}{2}(\frac{\partial z}{\partial a}+\frac{\partial z}{\partial b})$$ and $$\frac{\partial z}{\partial y} = \frac{\partial z}{\partial a}\frac{1}{2}+\frac{\partial z}{\partial b}(-\frac{1}{2})=\frac{1}{2}(\frac{\partial z}{\partial a}-\frac{\partial z}{\partial b})$$

Therefore $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{1}{4}\{(\frac{\partial z}{\partial a}+\frac{\partial z}{\partial b})^2-(\frac{\partial z}{\partial a}-\frac{\partial z}{\partial b})^2\}=\frac{1}{4}(4\frac{\partial z}{\partial a}\frac{\partial z}{\partial b})=\frac{\partial z}{\partial a} \frac{\partial z}{\partial b}$$ Hence $$\left(\frac{\partial z}{\partial x}\right)^{2}-\left(\frac{\partial z}{\partial y}\right)^{2}=\frac{\partial z}{\partial a} \frac{\partial z}{\partial b}$$