Question

If $u={\tan }^{-1}\left(\frac{x^6+y^6}{x^2+y^2}\right)$ then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=$

• A. $4\tan u$
• B. $4\sin u$
• C. $2\sin 2u$
• D. $2\tan 2u$

Answer 1

The correct option is C $2\sin 2u$

$a^3+b^3=(a+b)(a^2-ab+b^2)$

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

$=\frac{(x^2+y^2)(x^4-x^2{y}^2+y^4)}{(x^2+y^2)}$

$\tan u=x^4-x^2{y}^2+y^4$

${\sec }^{2}u\frac{du}{dx}=4x^3-2x{y}^2$

${\sec }^{2}u\frac{du}{dy}=4y^3-2x^{2}y$

${\sec }^{2}u(x\frac{du}{dx}+y\frac{du}{dy})=4x^4-2x^2{y}^2+4y^4-2x^2{y}^2$

$=4(x^4+y^4-x^2{y}^2)$

$x\frac{du}{dx}+y\frac{du}{dy}=4\tan u{\cos }^{2}u$

$=2\sin 2u$