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If $(x+y) \sin u=x^{2} y^{2}$, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=$
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Verified Answer
The correct answer is:
$\tan \mathrm{u}$
Given : $(\mathrm{x}+\mathrm{y}) \sin \mathrm{U}=\mathrm{x}^{2} \mathrm{y}^{2}$
$\Rightarrow \sin \mathrm{U}=\frac{\mathrm{x}^{2} \mathrm{y}^{2}}{\mathrm{x}+\mathrm{y}}=\mathrm{v}(\mathrm{let})$
Here $\mathrm{n}=2-1=1$
Euler's theorem $\mathrm{x} \cdot \frac{\partial \mathrm{v}}{\partial \mathrm{x}}+\mathrm{y} \cdot \frac{\partial \mathrm{v}}{\partial \mathrm{y}}=\mathrm{nv}$
$\therefore \mathrm{x} \frac{\partial \sin \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \sin \mathrm{U}}{\partial \mathrm{y}}=\sin \mathrm{U}$
$\Rightarrow \mathrm{x} \cdot \cos \mathrm{U} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \cdot \cos \mathrm{U} \cdot \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\sin \mathrm{U}$
$\Rightarrow \mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\frac{\sin \mathrm{U}}{\cos \mathrm{U}}=\tan \mathrm{U}$
$\Rightarrow \mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\tan \mathrm{U}$
$\Rightarrow \sin \mathrm{U}=\frac{\mathrm{x}^{2} \mathrm{y}^{2}}{\mathrm{x}+\mathrm{y}}=\mathrm{v}(\mathrm{let})$
Here $\mathrm{n}=2-1=1$
Euler's theorem $\mathrm{x} \cdot \frac{\partial \mathrm{v}}{\partial \mathrm{x}}+\mathrm{y} \cdot \frac{\partial \mathrm{v}}{\partial \mathrm{y}}=\mathrm{nv}$
$\therefore \mathrm{x} \frac{\partial \sin \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \sin \mathrm{U}}{\partial \mathrm{y}}=\sin \mathrm{U}$
$\Rightarrow \mathrm{x} \cdot \cos \mathrm{U} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \cdot \cos \mathrm{U} \cdot \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\sin \mathrm{U}$
$\Rightarrow \mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\frac{\sin \mathrm{U}}{\cos \mathrm{U}}=\tan \mathrm{U}$
$\Rightarrow \mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\tan \mathrm{U}$
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