Question

If $u={\tan }^{-1}\left(\frac{x^3+y^3}{x-y}\right)$, then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=$

• A. $\sin 2u$
• B. $\cos 2u$
• C. $\tan 2u$
• D. $\sec 2u$

Answer 1

The correct option is A $\sin 2u$

$\tan u=\frac{x^3+y^3}{(x-y)}$
${\sec }^{2}u\frac{\partial u}{\partial x}=\frac{3x^2(x-y)-(x^3+y^3)}{(x-y{)}^2}$
${\sec }^{2}u\frac{\partial u}{\partial y}=\frac{3y^2(x-y)-(x^3+y^3)}{(x-y{)}^2}$
${\sec }^{2}u(x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y})=\frac{3(x^3+y^3)(x-y)-(x-y)(x^3+y^3)}{(x-y{)}^2}=2\tan u$
$x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=\sin 2u$