Question
Find $\frac{d y}{d x}$ if, :
$
\sqrt{x}+\sqrt{y}=\sqrt{a}
$

Answer

$
\sqrt{x}+\sqrt{ } y=\sqrt{a}
$
Differentiating both sides w.r.t. $x_r$ we get
$
\begin{aligned}
& \frac{1}{2 \sqrt{x}}+\frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}=0 \\
& \therefore \frac{1}{2 \sqrt{y}} \cdot \frac{d y}{d x}=-\frac{1}{2 \sqrt{x}}
\end{aligned}
$
$
\therefore \frac{d y}{d x}=-\frac{2 \sqrt{y}}{2 \sqrt{x}}=-\sqrt{\frac{y}{x}} \text {. }
$

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