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Question Number 176374 by mathocean1 last updated on 17/Sep/22

$$Show:$$$$\frac{\partial X}{\partial Y}{\mid }_Z\frac{\partial Y}{\partial Z}{\mid }_X\frac{\partial X}{\partial Z}{\mid }_Y=-1$$

Answered by aleks041103 last updated on 17/Sep/22

$$dX=\frac{\partial X}{\partial Y}dY+\frac{\partial X}{\partial Z}dZ$$$$dY=\frac{\partial Y}{\partial Z}dZ+\frac{\partial Y}{\partial X}dX$$$$dZ=\frac{\partial Z}{\partial X}dX+\frac{\partial Z}{\partial Y}dY$$$$\Rightarrow dX=\frac{\partial X}{\partial Y}(\frac{\partial Y}{\partial Z}dZ+\frac{\partial Y}{\partial X}dX)+\frac{\partial X}{\partial Z}dZ=$$$$=(\frac{\partial X}{\partial Y}\frac{\partial Y}{\partial Z}+\frac{\partial X}{\partial Z})dZ+\frac{\partial X}{\partial Y}\frac{\partial Y}{\partial X}dX$$$$But\frac{\partial X}{\partial Y}=\frac{1}{\frac{\partial Y}{\partial X}}\Rightarrow \frac{\partial X}{\partial Y}\frac{\partial Y}{\partial X}=1$$$$\Rightarrow (1-1)dX=(\frac{\partial X}{\partial Y}\frac{\partial Y}{\partial Z}+\frac{\partial X}{\partial Z})dZ=0$$$$\Rightarrow \frac{\partial X}{\partial Y}\frac{\partial Y}{\partial Z}+\frac{\partial X}{\partial Z}=0$$$$\Rightarrow \frac{\partial X}{\partial Y}\frac{\partial Y}{\partial Z}=-\frac{\partial X}{\partial Z}$$$$\Rightarrow \frac{\partial X}{\partial Y}\frac{\partial Y}{\partial Z}\frac{\partial Z}{\partial X}=-1$$

Commented by mathocean1 last updated on 01/Oct/22

$$thanks$$

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