Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 12

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

$\displaystyle \frac{d}{dx} (1) + \frac{d}{dx} (x) = \frac{d}{dx} \left[ \sin (xy^2)\right]$


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



$
\begin{equation}
\begin{aligned}
1 &= \cos (xy^2) (2xyy' + y^2)\\
\\
1 &= 2xyy' \cos(xy^2) + y^2 \cos (xy^2)\\
\\
2xyy' \cos(xy^2) &= 1 - y^2 \cos (xy^2)\\
\\
\frac{y' \cancel{(2xy \cos(xy^2))}}{\cancel{2xy \cos(xy^2)}} &= \frac{1-y^2 \cos (xy^2)}{2xy \cos (xy^2)}\\
\\
y' &= \frac{1-y^2 \cos (xy^2)}{2xy \cos (xy^2)}
\end{aligned}
\end{equation}
$