Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 17

Determine $\displaystyle \frac{dy}{dx}$ of $\sqrt{xy} = 1 + x^2y$ by Implicit Differentiation.

$\displaystyle \frac{d}{dx} (\sqrt{xy}) = \frac{d}{dx} (1) \frac{d}{dx} (x^2y)$


$
\begin{equation}
\begin{aligned}
\frac{d}{dx} (xy)^{\frac{1}{2}} &= \frac{d}{dx} (1) + \left[ (x^2) \frac{d}{dx} (y) + (y) \frac{d}{dx} (x^2) \right]\\
\\
\frac{1}{2} (xy)^{\frac{-1}{2}} \cdot \frac{d}{dx} (xy) &= 0 + x^2 \frac{dy}{dx} + (y) (2x)\\
\\
\left( \frac{1}{2\sqrt{xy}}\right) \left[ (x) \frac{d}{dx} (y) + (y) \frac{d}{dx} (x) \right] &= x^2 \frac{dy}{dx} + 2xy\\
\\
\left( \frac{1}{2\sqrt{xy}}\right) \left[ x \frac{dy}{dx} + (y)(1) \right] &= x^2 \frac{dy}{dx} + 2xy
\end{aligned}
\end{equation}
$



$
\begin{equation}
\begin{aligned}
\left( \frac{1}{2\sqrt{xy}}\right) (xy'+y) &= x^2y' + 2xy\\
\\
\frac{xy'+y}{2\sqrt{xy}} & = x^2y' + 2xy\\
\\
xy'+y &= 2 \sqrt{xy} \left( x^2y' + 2xy\right)\\
\\
xy'+y &= 2x^2y' \sqrt{xy} + 4xy \sqrt{xy}\\
\\
xy' - 2x^2y' \sqrt{xy} &= 4xy \sqrt{xy} - y\\
\\
y'\left(x-2x^2 \sqrt{xy}\right) &= 4xy \sqrt{xy} - y\\
\\
\frac{y'\cancel{\left(x-2x^2\sqrt{xy}\right)}}{\cancel{x-2x^2\sqrt{xy}}} &= \frac{4xy \sqrt{xy}-y}{x-2x^2\sqrt{xy}}\\
\\
y' &= \frac{4xy \sqrt{xy}-y}{x-2x^2\sqrt{xy}}
\end{aligned}
\end{equation}
$