Question

If $x\sqrt{y}+y\sqrt{x}=1$, then $\frac{dy}{dx}$ equals

• A. $-\frac{y+2\sqrt{xy}}{x+2\sqrt{xy}}$
• B. $-\sqrt{\frac{x}{y}}\left(\frac{y+2\sqrt{xy}}{x+2\sqrt{xy}}\right)$
• C. $-\sqrt{\frac{y}{x}}\left(\frac{y+2\sqrt{xy}}{x+2\sqrt{xy}}\right)$
• D. None of these

Answer 1

Answer: (C)

$-\sqrt{\frac{y}{x}}\left(\frac{y+2\sqrt{xy}}{x+2\sqrt{xy}}\right)$

$x\sqrt{y}+y\sqrt{x}=1$

$\Rightarrow \sqrt{y}+\frac{x}{2\sqrt{y}}\frac{dy}{dx}+\frac{dy}{dx}\sqrt{x}+y\frac{1}{2\sqrt{x}}=0$

$\Rightarrow (\frac{x}{2\sqrt{y}}+\sqrt{x})\frac{dy}{dx}=-\frac{y}{2\sqrt{x}}-\sqrt{y}$

$\Rightarrow \left(\frac{x+2\sqrt{xy}}{2\sqrt{y}}\right)\frac{dy}{dx}=-\frac{y-2\sqrt{xy}}{2\sqrt{x}}$

$\Rightarrow \frac{dy}{dx}=-\sqrt{\frac{y}{x}}\left(\frac{y+2\sqrt{xy}}{x+2\sqrt{xy}}\right)$