Question

$I$f $u={\sin }^{-1}\left(\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}\right)$

then $\frac{\partial u}{\partial x}=?$

• A. $\frac{y}{x}\frac{\partial u}{\partial y}$
• B. $-\frac{y}{x}\frac{\partial u}{\partial y}$
• C. $\frac{x}{y}\frac{\partial u}{\partial y}$
• D. $\frac{-x}{y}\frac{\partial u}{\partial y}$

Answer 1

The correct option is B $-\frac{y}{x}\frac{\partial u}{\partial y}$

$sinu=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}+\sqrt{y}}$

$cosu\frac{du}{dx}=\frac{\frac{1}{2}\frac{1}{\sqrt{x}}(\sqrt{x}+\sqrt{y})-\frac{1}{2\sqrt{x}}(\sqrt{x}-\sqrt{y})}{(\sqrt{x}+\sqrt{y}{)}^2}$

$cosu\frac{du}{dx}=\frac{\sqrt{\frac{y}{x}}}{(\sqrt{x}+\sqrt{y}{)}^2}$..........(1)

$cosu\frac{du}{dy}=\frac{-\sqrt{\frac{x}{y}}}{(\sqrt{x}+\sqrt{y}{)}^2}$..........(2)

$\frac{du}{dx}=-\frac{y}{x}\frac{du}{dy}$