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If $\dfrac{x^2}{1+u}+\dfrac{y^2}{2+u}+\dfrac{z^2}{3+u}=1$ then prove the following

Prove : $\bigg(\dfrac{\partial u}{\partial x} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial y} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\; 2 \bigg( x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} \bigg) \\ \; \\$

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Now, $\dfrac{x^2}{1+u}+\dfrac{y^2}{2+u}+\dfrac{z^2}{3+u}=1$

Differentiating given equation partially w.r.t. x,

$\therefore \dfrac{2x}{2+u} - \dfrac{x^2}{(1+u)^2} \dfrac{\partial u}{\partial x} - \dfrac{y^2}{(2+u)^2} \dfrac{\partial u}{\partial x} - \dfrac{z^2}{(3+u)^2} \dfrac{\partial u}{\partial x} \\ \; \\ \; \\ \therefore \dfrac{2x}{1+u} \;=\; \dfrac{\partial u}{\partial x} \bigg[ \dfrac{x^2}{(1+u)^2}+ \dfrac{y^2}{(2+u)^2}+ \dfrac{z^2}{(3+u)^2} \bigg] \\ \; \\ \; \\ \therefore \dfrac{2x}{1+u} \;=\; p . \dfrac{\partial u}{\partial x} \; where \; \; p\;=\; \dfrac{x^2}{(1+u)^2}+ \dfrac{y^2}{(2+u)^2}+ \dfrac{z^2}{(3+u)^2} \; \; \; \ldots (i) \\ \; \\ \; \\ \therefore \dfrac{\partial u}{\partial x} \;=\; \dfrac{2x}{(1+u) p}$

Similarly, we can find out

$\therefore \dfrac{\partial u}{\partial y} \;=\; \dfrac{2y}{(1+u) p} \; and \; \dfrac{\partial u}{\partial z} \;=\; \dfrac{2z}{(1+u) p} \\ \; \\ \; \\ \bigg(\dfrac{\partial u}{\partial x} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial y} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\; \dfrac{4x^2}{(1+u)^2 . p} + \dfrac{4y^2}{(2+u)^2 . p} + \dfrac{4z^2}{(3+u)^2 . p} \\ \; \\ \; \\ = \dfrac{4}{p^2} \bigg[ \dfrac{x^2}{(1+u)^2}+ \dfrac{y^2}{(2+u)^2}+ \dfrac{z^2}{(3+u)^2} \bigg] \\ \; \\ \; \\ = \dfrac{4}{p^2} . p \\ \; \\ \; \\ \bigg(\dfrac{\partial u}{\partial x} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial y} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\; \dfrac{4}{p} \; \; \; \; \; \ldots (ii)$

Now, $\therefore x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} \;=\; \dfrac{2x^2}{(1+u). p} + \dfrac{2y^2}{(2+u). p} + \dfrac{2z^2}{(3+u). p} \\ \; \\ \; \\ \therefore x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} \;=\; \dfrac{2}{p} \Bigg[ \dfrac{x^2}{(1+u)} + \dfrac{y^2}{(2+u)} + \dfrac{z^2}{(3+u)} \Bigg] \\ \; \\ \; \\ = \dfrac{2}{p} (1) \\ \; \\ \; \\ \therefore x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} \;=\; \dfrac{2}{p}$

From (ii) and (iii), we can prove that

$\bigg(\dfrac{\partial u}{\partial x} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial y} \bigg)^2 + \bigg(\dfrac{\partial u}{\partial z} \bigg)^2 \;=\; 2 \bigg( x\dfrac{\partial u}{\partial x} + y\dfrac{\partial u}{\partial y} + z \dfrac{\partial u}{\partial z} \bigg)$

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