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
$x\frac{\partial f}{\partial x}+y\frac{\partial f}{\partial y}=f$
$\Rightarrow x\frac{\partial }{\partial x}(\sin u)+y\frac{\partial }{\partial y}(\sin u)=\sin u$
$\Rightarrow x\cos u\frac{\partial u}{\partial x}+y\cos u\frac{\partial u}{\partial y}=\sin u$
$\Rightarrow x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\tan u$