Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 15

Determine $\displaystyle \frac{dy}{dx}$ of $\tan \left( \frac{x}{y} \right) = x + y$ by Implicit Differentiation.

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

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



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