MCQ
જો $u = {\tan ^{ - 1}}(x + y),$ તો $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${\sec ^2}u$
$\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 $ $x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = \tan u.{\cos ^2}u = \sin u\cos u$ = $\frac{1}{2}\sin 2u$.
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