Q110 / 160 · TS EAMCET 2006 · Mathematics

If $u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ equal to :

  • A
    $\sin u$
  • Correct answer
    B
    $\tan u$
  • C
    $\cos u$
  • D
    $\cot u$

Solution

$u=\sin ^{-1}\left(\frac{x^2+y^2}{x+y}\right)$ Here $u$ is not a homogeneous function. But $f(x, y)=\sin u=\frac{x^2+y^2}{x+y}$ is a homogeneous function of degree one. Here by Euler's theorem $x \frac{\partial f}{\partial x}+y \frac{\partial f}{\partial y}=f$ $\Rightarrow \quad x \frac{\partial}{\partial x}(\sin u)+y \frac{\partial}{\partial y}(\sin u)=\sin u$ $\Rightarrow \quad x \cos u \frac{\partial u}{\partial x}+y \cos u \frac{\partial u}{\partial y}=\sin u$ $\Rightarrow \quad x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\tan u$

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

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