Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 16

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

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

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



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