If $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$, then $\frac{d y}{d x}$ is equal to
If $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$, then $\frac{d y}{d x}$ is equal to
$\frac{x^2+y^2}{x+y}$
$\frac{x^2-y^2}{x+y}$
$1$
$2$
Solution
We have $\sqrt{\frac{y}{x}}+\sqrt{\frac{x}{y}}=2$
$
\Rightarrow \quad y+x=2 \sqrt{x} \cdot \sqrt{y}
$
On squaring both sides, we get
$
\begin{array}{cl}
& x^2+y^2+2 x y=4 x y \\
\Rightarrow & x^2+y^2-2 x y=0 \\
\Rightarrow & (x-y)^2=0 \Rightarrow y=x \\
\therefore & \frac{d y}{d x}=1
\end{array}
$