Question

If (x +y )sin u = $x^2y^2$, then $x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} =$

• A. sin u
• B. cosec u
• C. 2 tan u
• D. tan u

Answer 1

Answer: (D)

tan u

Answer 2

Given : (x + y) sin U = $x^2y^2$
$\left(x+y\right)\sin U =x^{2}y^{2}$
$\Rightarrow \sin U = \frac{x^{2}y^{2}}{x+y} =v$ (let)
Here n = 2 - 1 = 1
Euler's theorem $x. \frac{\partial v}{\partial x} +y. \frac{\partial v}{\partial y} =nv$
$\therefore x \frac{\partial\sin U}{\partial x} + y \frac{\partial\sin U}{\partial y} =\sin U$
$\Rightarrow x.\cos U \frac{\partial U}{\partial x} +y .\cos U. \frac{\partial U}{\partial y} =\sin U$
$\Rightarrow x \frac{\partial U}{\partial x} +y \frac{\partial U}{\partial y} = \frac{\sin U}{\cos U} =\tan U$
$\Rightarrow x \frac{\partial U}{\partial x} + y \frac{\partial U}{\partial y} =\tan U$