MCQ
If $u = {\tan ^{ - 1}}\left( {{{{x^3} + {y^3}} \over {x - y}}} \right)$, then $x{{\partial u} \over {\partial x}} + y{{\partial u} \over {\partial y}} = $
- ✓$\sin 2u$
- B$\cos 2u$
- C$\tan 2u$
- D$\sec 2u$
$\therefore $ $x\frac{\partial }{{\partial x}}(\tan u) + y\frac{\partial }{{\partial y}}(\tan u) = 2(\tan u)$
$\therefore $ $x{\sec ^2}u\frac{{\partial u}}{{\partial x}} + y{\sec ^2}u\frac{{\partial u}}{{\partial y}} = 2\tan u$
==> $x\frac{{\partial u}}{{\partial x}} + y\frac{{\partial u}}{{\partial y}} = 2\frac{{\tan u}}{{{{\sec }^2}u}}$ = $2\sin u\cos u = \sin 2u$.
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$ =\frac{1}{2024}, $ then $\alpha$ is equal to-