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If $u=x y^2 \tan ^{-1}\left(\frac{y}{x}\right)$, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to
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$3 u$
Given that,
$\begin{aligned} & u=x y^2 \tan ^{-1}\left(\frac{y}{x}\right) \\ & \frac{\partial u}{\partial x}=y^2 \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(-\frac{y}{x^2}\right) \\ &=y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x y^3}{x^2+y^2} \\ & \frac{\partial u}{\partial y}=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(\frac{1}{x}\right) \\ &=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+\frac{x^2 y^2}{x^2+y^2} \\ & \therefore \quad x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y}=x y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x^2 y^3}{x^2+y^2} \\ & x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 x y^2 \tan ^{-1} \frac{y}{x}=3 u\end{aligned}$
$\begin{aligned} & u=x y^2 \tan ^{-1}\left(\frac{y}{x}\right) \\ & \frac{\partial u}{\partial x}=y^2 \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(-\frac{y}{x^2}\right) \\ &=y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x y^3}{x^2+y^2} \\ & \frac{\partial u}{\partial y}=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+x y^2 \cdot \frac{1}{1+\frac{y^2}{x^2}} \cdot\left(\frac{1}{x}\right) \\ &=2 x y \tan ^{-1}\left(\frac{y}{x}\right)+\frac{x^2 y^2}{x^2+y^2} \\ & \therefore \quad x \frac{\partial u}{\partial x}+ y \frac{\partial u}{\partial y}=x y^2 \tan ^{-1}\left(\frac{y}{x}\right)-\frac{x^2 y^3}{x^2+y^2} \\ & x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 x y^2 \tan ^{-1} \frac{y}{x}=3 u\end{aligned}$
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