Q.2 (a) If \(H=f(y-z,\,z-x,\,x-y)\), then prove that \(\frac{\partial H}{\partial x}+\frac{\partial H}{\partial y}+\frac{\partial H}{\partial z}=0\).
Best answer
Let \(u=y-z,\ v=z-x,\ w=x-y\), so \(H=f(u,v,w)\).
\(\frac{\partial u}{\partial x}=0,\ \frac{\partial v}{\partial x}=-1,\ \frac{\partial w}{\partial x}=1\); \(\frac{\partial u}{\partial y}=1,\ \frac{\partial v}{\partial y}=0,\ \frac{\partial w}{\partial y}=-1\); \(\frac{\partial u}{\partial z}=-1,\ \frac{\partial v}{\partial z}=1,\ \frac{\partial w}{\partial z}=0\).
By the chain rule:
\(\frac{\partial H}{\partial x}=-\frac{\partial H}{\partial v}+\frac{\partial H}{\partial w}\)
\(\frac{\partial H}{\partial y}=\frac{\partial H}{\partial u}-\frac{\partial H}{\partial w}\)
\(\frac{\partial H}{\partial z}=-\frac{\partial H}{\partial u}+\frac{\partial H}{\partial v}\)
Adding: \(\frac{\partial H}{\partial x}+\frac{\partial H}{\partial y}+\frac{\partial H}{\partial z}=0\) (proved).
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