If $u=\sin ^{-1}\left(\frac{x}{y}\right)+\tan ^{-1}\left(\frac{y}{x}\right)$, then the value $x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}$ is

• Correct answer
  A

  0
• B

  1
• C

  2
• D

  None of these

Solution

$\begin{aligned} & \because u=\sin ^{-1}\left(\frac{x}{y}\right)+\tan ^{-1}\left(\frac{y}{x}\right) \\ & \therefore \frac{\partial u}{\partial x}=\frac{1}{\sqrt{\left\{1-\left(\frac{x}{y}\right)^2\right\}}} \frac{1}{y}+\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot\left(-\frac{y}{x^2}\right)\end{aligned}$ $=\frac{1}{\sqrt{\left(y^2-x^2\right)}}-\frac{y}{\left(x^2+y^2\right)}$
$\begin{array}{r}\frac{\partial u}{\partial y}=\frac{1}{\sqrt{\left\{1-\left(\frac{x}{y}\right)^2\right\}}} \cdot\left(-\frac{x}{y^2}\right)+\frac{1}{1+\left(\frac{y}{x}\right)^2} \cdot\left(\frac{1}{x}\right) \\ =-\frac{x}{y \sqrt{\left(y^2-x^2\right)}}+\frac{x}{\left(x^2+y^2\right)}\end{array}$
On adding Eqs. (i) and (ii), we get $$ x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=0 $$

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