Question

If $u={\tan }^{-1}\left(\frac{x^2+y^2}{x+y}\right)$, then $x\frac{du}{dx}+y\frac{du}{dy}=$

• A. $\sin 2u$
• B. $\frac{1}{2}\sin 2u$
• C. $\frac{1}{3}\sin 2u$
• D. $2\sin 2u$

Answer 1

Answer: (B)

$\frac{1}{2}\sin 2u$

Given, $u={\tan }^{-1}\left(\frac{x^2+y^2}{x+y}\right)$

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

On differentiating both the sides w.r.t. $x$ and $y$ respectively, we get

${\sec }^{2}u\frac{du}{dx}=\frac{2x(x+y)-(x^2+y^2)}{(x+y{)}^2}$ ....(1)

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

Multiplying $x$ with equation (1) and $y$ to equation (2).

Therefore, $x{\sec }^{2}u\frac{du}{dx}=\frac{x^3+2x^{2}y-x{y}^2}{(x+y{)}^2}$ ....(3)

and $y{\sec }^{2}u\frac{du}{dy}=\frac{y^3+2x{y}^2-x^{2}y}{(x+y{)}^2}$ ....(4)

Adding equations (3) and (4), we get

${\sec }^{2}u(x\frac{du}{dx}+y\frac{du}{dy})=\frac{x^3+2x^{2}y-x{y}^2+y^3+2x{y}^2-x^{2}y}{(x+y{)}^2}$

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

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

$=\frac{x^2+y^2}{x+y}$

$=\tan u$

Thus $x\frac{du}{dx}+y\frac{du}{dy}=\sin u\cos u$

$=\frac{1}{2}\sin 2u$