If $z = f (u, v), u = x^{2} - 2 x y - y^{2}, v = a$ , then

(x + y) \frac{\partial z}{\partial x} = (x - y) \frac{\partial z}{\partial y}

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(x - y) \frac{\partial z}{\partial x} = (x + y) \frac{\partial z}{\partial y}

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(x + y) \frac{\partial z}{\partial x} = (y - x) \frac{\partial z}{\partial y}

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(y - x) \frac{\partial z}{\partial x} = (x - y) \frac{\partial z}{\partial y}

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Solution

The correct option is C $(x + y) \frac{\partial z}{\partial x} = (y - x) \frac{\partial z}{\partial y}$
Given, $z = f (u, v); u = x^{2} - 2 x y - y^{2}, v = a$
$\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} . \frac{\partial u}{\partial x} + \frac{\partial z}{\partial v} . \frac{\partial v}{\partial x}$
and $\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} . \frac{\partial u}{\partial y} + \frac{\partial z}{\partial v} . \frac{\partial v}{\partial y}$
Also, $\frac{\partial u}{\partial x} = 2 x - 2 y$ $\frac{\partial u}{\partial y} = - 2 x - 2 y$
$\frac{\partial v}{\partial x} = 0$ $\frac{\partial v}{\partial y} = 0$

$\frac{\partial z}{\partial x} = (2 x - 2 y) (\frac{\partial z}{\partial u}) \dots (i)$ $\frac{\partial z}{\partial y} = - (2 x + 2 y) (\frac{\partial z}{\partial u}) \dots (i i)$
From equation (i) and (ii), we get
$(x + y) \frac{\partial z}{\partial x} = (y - x) \frac{\partial z}{\partial y}$
Hence, option (c) is correct.

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