Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 18

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

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


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

\end{aligned}
\end{equation}
$




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