Single Variable Calculus, Chapter 7, 7.6, Section 7.6, Problem 38

If $\tan^{-1} (xy) = 1 + x^2 y$. Find $y'$
If we take the derivative implicity, we have...


$
\begin{equation}
\begin{aligned}
\frac{\frac{d}{dx}(xy)}{1 + (xy)^2} &= 0 + \left[ x^2 \cdot \frac{d}{dy} (y) \frac{dy}{dx} + 2x \cdot y \right]\\
\\
\frac{x \frac{dy}{dx}+y(1)}{1+(xy)^2} &= x^2 \frac{dy}{dx} + 2xy\\
\\
x \frac{dy}{dx} + y &= \left[ 1 + (xy)^2 \right] \left( x^2 \frac{dy}{dx} + 2xy \right)\\
\\
x \frac{dy}{dx} \left[ 1 + (xy)^2 \right] \left( x^2 \frac{dy}{dx} \right) &= y\left( 2x + 2x^3 y^2 - 1\right)\\
\\
\frac{dy}{dx} &= \frac{y\left( 2x + 2x^3 y^2 - 1\right)}{x\left( 1 - x \left[ 1 + (xy)^2 \right] \right)}\\
\\
\frac{dy}{dx} &= \frac{y\left( 2x + 2x^3 y^2 - 1\right)}{x\left( 1 - x - x^3y^2 \right)}
\end{aligned}
\end{equation}
$