Question

If u=f(x-y,y-z,z-x) prove that du/dx+du/dy+du/dz=0

Answer 1

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Answer 2

Given,

u=f(x-y,y-z,z-x)

To Find,

prove that $\frac{du}{dx} +\frac{du}{dy}+\frac{du}{dz} =0$

Solution,

suppose X= y-z, Y= z-x, Z= x-y, then u= f(X,Y,Z)

therefore u is composite function x,y,z respectively.

we have, X= y-z, Y= z-x, Z= x-y

partially differentiate w.r.t x,y,z respectively, we get

$\frac{dX}{dx}=0, \frac{dX}{dy}= 1, \frac{dX}{dz}=-1$

and $\frac{dY}{dx}=-1, \frac{dY}{dy}= 0, \frac{dY}{dz}=1$

and $\frac{dZ}{dx}=1, \frac{dZ}{dy}= -1, \frac{dZ}{dz}=0$

now, $\frac{du}{dx} =\frac{du}{dX} .\frac{dX}{dx} +\frac{du}{dY} .\frac{dY}{dx} +\frac{du}{dZ} .\frac{dZ}{dx}$

             = $\frac{du}{dX} .(0) +\frac{du}{dY} .(-1) +\frac{du}{dZ} .(1) =- \frac{du}{dY} +\frac{du}{dZ}$              (1)

        $\frac{du}{dy} =\frac{du}{dX} .\frac{dX}{dy} +\frac{du}{dY} .\frac{dY}{dy} +\frac{du}{dZ} .\frac{dZ}{dy}$                                    

             =  $\frac{du}{dX} .(1) +\frac{du}{dY} .(0) +\frac{du}{dZ} .(-1) =\frac{du}{dX} -\frac{du}{dZ}$                (2)

        $\frac{du}{dZ} =\frac{du}{dX} .\frac{dX}{dz} +\frac{du}{dY} .\frac{dY}{dz} +\frac{du}{dZ} .\frac{dZ}{dz}$

             = $\frac{du}{dX} .(-1) +\frac{du}{dY} .(1) +\frac{du}{dZ} .(0) =-\frac{du}{dX} +\frac{du}{dY}$               (3)

adding (1),(2),(3) we get,

$\frac{du}{dx} +\frac{du}{dy}+\frac{du}{dz}$ $= - \frac{du}{dY} +\frac{du}{dZ}$$+\frac{du}{dX} -\frac{du}{dZ}$$-\frac{du}{dX} +\frac{du}{dY}$ =0

Hence $\frac{du}{dx} +\frac{du}{dy}+\frac{du}{dz}$ = 0.