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. $cotu$

Answer 1

The correct option is B

$\tan u$

Explanation for the correct answer:

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

Given,

$\begin{array}{rcl}u& =& {\sin }^{-1}\left(\frac{(x^2+y^2)}{(x+y)}\right)\\ \sin u& =& \frac{(x^2+y^2)}{(x+y)}\\ & =& \frac{x^2\left(\frac{1+y^2}{x^2}\right)}{x\left({\displaystyle \frac{1+y}{x}}\right)}\\ & =& \frac{x\left(\frac{1+y^2}{x^2}\right)}{\left({\displaystyle \frac{1+y}{x}}\right)}\\ & =& xf\left(\frac{y}{x}\right)\end{array}$

This is a homogeneous function of the degree$1.$

So by Euler’s theorem,

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}=nz$

Let, $z=\sin u$

Then,

$\begin{array}{rcl}x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y}& =& z\\ & \Rightarrow & x\left(\frac{\partial }{\partial x}\right)\sin u+y\left(\frac{\partial }{\partial y}\right)\sin u=\sin u\\ & \Rightarrow & x\cos u\left(\frac{\partial u}{\partial x}\right)+y\cos u\left(\frac{\partial u}{\partial x}\right)=\sin u\left[\because \frac{\partial \left(sinu\right)}{\partial x}=cosu\right]\mathbf{}\\ & \Rightarrow & x\left(\frac{\partial u}{\partial x}\right)+y\left(\frac{\partial u}{\partial y}\right)=\frac{\sin u}{\cos u}\\ & \Rightarrow & x\left(\frac{\partial u}{\partial x}\right)+y\left(\frac{\partial u}{\partial y}\right)=\tan u\end{array}$

Hence the correct answer is option (B).