Question

If $u={\sin }^{-1}\left(\frac{x^2+y^2}{x+y}\right)$

then $x\frac{\partial u}{\partial x}+y\frac{\partial u}{\partial y}=$

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

Answer 1

Answer: (D)

$\tan u$

Given $\sin u=\frac{x^2+y^2}{x+y}$

Differentiating w.r.t. x, we get,

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

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

$x\frac{du}{dx}+y\frac{du}{dy}=secu\left(\frac{y^3+y^{2}x+x^{2}y+x^3}{(x+y{)}^2}\right)$

$=\sec u\sin u$

$=tanu$