Q.2 (c) If \(u=f\left(e^{x-y},e^{y-z},e^{z-x}\right)\), then prove that \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=0\).
Best answer
Let \(X=e^{x-y},\ Y=e^{y-z},\ Z=e^{z-x}\), so \(u=f(X,Y,Z)\).
\(\frac{\partial X}{\partial x}=X,\ \frac{\partial X}{\partial y}=-X,\ \frac{\partial X}{\partial z}=0\); \(\frac{\partial Y}{\partial x}=0,\ \frac{\partial Y}{\partial y}=Y,\ \frac{\partial Y}{\partial z}=-Y\); \(\frac{\partial Z}{\partial x}=-Z,\ \frac{\partial Z}{\partial y}=0,\ \frac{\partial Z}{\partial z}=Z\).
Chain rule:
\(\frac{\partial u}{\partial x}=X\frac{\partial u}{\partial X}-Z\frac{\partial u}{\partial Z}\)
\(\frac{\partial u}{\partial y}=-X\frac{\partial u}{\partial X}+Y\frac{\partial u}{\partial Y}\)
\(\frac{\partial u}{\partial z}=-Y\frac{\partial u}{\partial Y}+Z\frac{\partial u}{\partial Z}\)
Adding: \(\frac{\partial u}{\partial x}+\frac{\partial u}{\partial y}+\frac{\partial u}{\partial z}=0\) (proved).
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