Question

$I$f $u=x{y}^2{\tan }^{-1}\left(\frac{y}{x}\right)$ then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}$

• A. $2u$
• B. $u$
• C. $3u$
• D. $\frac{1}{3}u$

Answer 1

The correct option is C $3u$

$u=x{y}^2{\tan }^{-1}\left(\frac{y}{x}\right)$

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

$\frac{\partial u}{\partial x}=y^2{\tan }^{-1}\left(\frac{y}{x}\right)-\frac{y^3}{\sqrt{x^2+y^2}}$ ..... $\left(i\right)$

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

$\frac{\partial u}{\partial y}=2xy{\tan }^{-1}\left(\frac{y}{x}\right)+\frac{x{y}^2}{\sqrt{x^2+y^2}}$ ..... $\left(ii\right)$

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

$=3x{y}^2{\tan }^{-1}\left(\frac{y}{x}\right)$
$=3u$