Question

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

• A. $0$
• B. $1$
• C. $2$
• D. none of these

Answer 1

The correct option is A

$0$

Explanation for the correct answer:

Find the value of $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$:

Given,

$\begin{array}{rcl}u& =& {\sin }^{-1}\left(\frac{x}{y}\right)+{\tan }^{-1}\left(\frac{y}{x}\right)\\ & =& {\sin }^{-1}\left(\frac{1}{{\displaystyle \frac{y}{x}}}\right)+{\tan }^{-1}\left(\frac{y}{x}\right)\\ & =& x^{0}f\left(\frac{y}{x}\right)\end{array}$

Here, $u$ is a homogeneous function of the degree $0.$

Here,$n=0$

So, by Euler’s theorem,

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=nu$

$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=0$

Hence, the correct option is A.