Q82 / 160 · TS EAMCET 2010 · Mathematics

If $u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$, then $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is equal to

  • A
    $3 u$
  • B
    $4 u$
  • C
    $3 \sin u$
  • Correct answer
    D
    $3 \tan u$

Solution

$u=\sin ^{-1}\left(\frac{x^4+y^4}{x+y}\right)$ Let $v=\sin u=\frac{x^4+y^4}{x+y}$, here degree is homogeneous, so $n=4-1=3$ By Euler's theorem, we have to prove that, $x \frac{\partial v}{\partial x}+y \frac{\partial v}{\partial y}=3 v$ $x \frac{\partial}{\partial x}(\sin u)+y \frac{\partial v}{\partial y}(\sin u)=3 \sin u$ $x \cos u \cdot \frac{\partial u}{\partial x}+y \cos u \frac{\partial u}{\partial y}=3 \sin u$ $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 \frac{\sin u}{\cos u}$ $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=3 \tan u$

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

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