MCQ
If $u = {\tan ^{ - 1}}(x + y),$ then $x{{\partial u} \over {\partial x}} + y{{\partial u} \over {\partial y}} = $
  • A
    $\sin 2u$
  • ${1 \over 2}\sin 2u$
  • C
    $2\tan u$
  • D
    ${\sec ^2}u$

Answer

Correct option: B.
${1 \over 2}\sin 2u$
b
(b) $\tan u = x + y = x.\left( {1 + \frac{y}{x}} \right)$

$\therefore $ $\tan u$ is homogeneous in $x,\,y$ of order $ 1$.

$\therefore $ $x\frac{\partial }{{\partial x}}(\tan u) + y\frac{\partial }{{\partial y}}(\tan u) = \tan u$

$\therefore $ $x{\sec ^2}u\frac{{\partial u}}{{\partial x}} + y{\sec ^2}u\frac{{\partial u}}{{\partial y}} = \tan u$

$\therefore $ $x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \tan u.{\cos ^2}u = \sin u\cos u$ = $\frac{1}{2}\sin 2u$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free