Q53 / 120 · VITEEE 2007 · Mathematics

If $(x+y) \sin u=x^{2} y^{2}$, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=$

  • A
    $\sin \mathrm{u}$
  • B
    $\operatorname{cosec} \mathrm{u}$
  • C
    $2 \tan \mathrm{u}$
  • Correct answer
    D
    $\tan \mathrm{u}$

Solution

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 $\mathrm{x} \cdot \frac{\partial \mathrm{v}}{\partial \mathrm{x}}+\mathrm{y} \cdot \frac{\partial \mathrm{v}}{\partial \mathrm{y}}=\mathrm{nv}$ $\therefore \mathrm{x} \frac{\partial \sin \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \sin \mathrm{U}}{\partial \mathrm{y}}=\sin \mathrm{U}$ $\Rightarrow \mathrm{x} \cdot \cos \mathrm{U} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \cdot \cos \mathrm{U} \cdot \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\sin \mathrm{U}$ $\Rightarrow \mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\frac{\sin \mathrm{U}}{\cos \mathrm{U}}=\tan \mathrm{U}$ $\Rightarrow \mathrm{x} \frac{\partial \mathrm{U}}{\partial \mathrm{x}}+\mathrm{y} \frac{\partial \mathrm{U}}{\partial \mathrm{y}}=\tan \mathrm{U}$

Hence, the correct answer is option: 'D'.

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