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If $u=f(x-y,y-z,z-x) ,$ then show that $\dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial u}{\partial z} \;=\; 0$
partial differentiation • 8.6k  views
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Let, p = x – y, q = y – z and r = z – x. ∴ u =f(p, q, r)

$\therefore \dfrac{\partial p}{\partial x}=1 \; \; and \; \; \dfrac{\partial p}{\partial y}=-1$

Similarly,

$\dfrac{\partial q}{\partial y}=1 \; , \; \dfrac{\partial q}{\partial z}=-1 \; , \; \dfrac{\partial r}{\partial z}=1 \; and \; \dfrac{\partial r}{\partial x}=-1 \; \; \\ \; \\ \; \\ Now , \; \dfrac{\partial u}{\partial x} \;=\; \dfrac{\partial u}{\partial p} . \dfrac{\partial p}{\partial x} + \dfrac{\partial u}{\partial r} . \dfrac{\partial r}{\partial x} \;=\; \dfrac{\partial u}{\partial p} (1)+ \dfrac{\partial u}{\partial r} (-1) \;=\; \dfrac{\partial u}{\partial p} - \dfrac{\partial u}{\partial r} \; \; \; \ldots (i)$

Similarly, $\dfrac{\partial u}{\partial y} \;=\; \dfrac{\partial u}{\partial p} . \dfrac{\partial p}{\partial y} + \dfrac{\partial u}{\partial q} . \dfrac{\partial q}{\partial y} \;=\; \dfrac{\partial u}{\partial p} (-1) + \dfrac{\partial u}{\partial q} (1) \;=\; \dfrac{\partial u}{\partial q} - \dfrac{\partial u}{\partial p} \; \; \; \ldots (ii)$

and $\dfrac{\partial u}{\partial z} \;=\; \dfrac{\partial u}{\partial q} . \dfrac{\partial q}{\partial z} + \dfrac{\partial u}{\partial r} . \dfrac{\partial r}{\partial z} \;=\; \dfrac{\partial u}{\partial q} (-1)+ \dfrac{\partial u}{\partial r} (1) \;=\; \dfrac{\partial u}{\partial r} - \dfrac{\partial u}{\partial q} \; \; \; \; \ldots (iii) \\ \; \\ \; \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial x} + \dfrac{\partial u}{\partial y} + \dfrac{\partial u}{\partial z} = \dfrac{\partial u}{\partial p} - \dfrac{\partial u}{\partial r} + \dfrac{\partial u}{\partial p} - \dfrac{\partial u}{\partial r} + \dfrac{\partial u}{\partial q} - \dfrac{\partial u}{\partial p} \; = \; 0$

Hence Proved.

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