If $\sqrt{x}+\sqrt{y}=\sqrt{x y}$, then $\frac{d y}{d x}=$
If $\sqrt{x}+\sqrt{y}=\sqrt{x y}$, then $\frac{d y}{d x}=$
$-\left(\frac{y}{x}\right)^{\frac{3}{2}}$
$\left(\frac{x}{y}\right)^{\frac{3}{2}}$
$-\left(\frac{x}{y}\right)^{\frac{3}{2}}$
$\left(\frac{y}{x}\right)^{\frac{3}{2}}$
Solution
We have $\sqrt{x}+\sqrt{y}=\sqrt{x y}$
Dividing both sides by $\sqrt{x y}$, we get
$\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{x}}=1$
Differentiating both sides w.r.t. $x$, we get
$\begin{aligned}
&\left(\frac{-1}{2}\right)\left(\frac{1}{y^{3 / 2}}\right)\left(\frac{d y}{d x}\right)-\left(\frac{1}{2}\right)\left(\frac{1}{x^{3 / 2}}\right)=0 \\
\therefore & \frac{d y}{d x}=-\left(\frac{y}{x}\right)^{3 / 2}
\end{aligned}$