Question

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

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

Answer 1

Answer: (B)

$u=\sin ^{-1}\left(\frac{x^{2}+y^{2}}{x+y}\right)$
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 $
$\Rightarrow x \frac{\partial}{\partial x}(\sin u)+y \frac{\partial}{\partial y}(\sin u)=\sin u$
$\Rightarrow x \cos u \frac{\partial u}{\partial x}+y \cos u \frac{\partial u}{\partial y}=\sin u $
$\Rightarrow x \frac{\partial u}{\partial x}+y \frac{\partial u}{\partial y}=\tan u$