Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 6

Determine $\displaystyle \frac{dy}{dx}$ of $2\sqrt{x}+\sqrt{y} =3 $ by Implicit Differentiation.

$\displaystyle \frac{d}{dx} (2\sqrt{x}) + \frac{d}{dx} (\sqrt{y}) = \frac{d}{dx} (3)$

$
\begin{equation}
\begin{aligned}
2 \frac{d}{dx} (x)^{\frac{1}{2}} + \frac{d}{dx} (y)^{\frac{1}{2}} = \frac{d}{dx} (3)\\
\\
(\cancel{2})\left(\frac{1}{\cancel{2}}\right)(x)^{\frac{-1}{2}} + \left( \frac{1}{2} \right) (y)^{\frac{-1}{2}} &= \frac{dy}{dx} (3)\\
\\
(x)^{\frac{-1}{2}} + \frac{1}{2}(y)^{\frac{-1}{2}} \frac{dy}{dx} &= 0\\
\\
\left( \frac{1}{2\sqrt{y}}\right) \frac{dy}{dx} &= -\frac{1}{\sqrt{x}}\\
\\
\frac{dy}{dx} &= \frac{-2\sqrt{y}}{x} \qquad \text{ or } \qquad y'= \frac{-2\sqrt{y}}{\sqrt{x}}
\end{aligned}
\end{equation}
$