Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 14

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

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


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



$
\begin{equation}
\begin{aligned}
2xy \cos (x^2) + y' \sin (x^2) &= 2xyy' \cos (y^2) + \sin (y^2)\\
\\
y' \sin (x^2) - 2xyy' \cos(y^2) &= \sin(y^2) - 2xy \cos (x^2)\\
\\
y'\left[ \sin(x^2) - 2xy \cos (y^2) \right] &= \sin (y^2) - 2 xy \cos (x^2)\\
\\
\frac{y' \cancel{\left[\sin (x^2) - 2xy \cos (y^2)\right]}}{\cancel{\sin (x^2) - 2xy \cos (y^2)}} &= \frac{\sin (y^2) - 2xy \cos (x^2)}{\sin (x^2) - 2xy \cos (y^2)}\\
\\
y' & = \frac{\sin (y^2) - 2xy \cos (x^2)}{\sin (x^2) - 2xy \cos (y^2)}
\end{aligned}
\end{equation}
$