Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 26

Find an equation of the tangent to the curve $x^2+2xy-y^2+x=2$ at the point $(1,2)$ using Implicit Differentiation.
If $y'= m\text{ (slope)}$ then,
$\displaystyle \frac{d}{dx} (x^2) + \frac{d}{dx} (2xy) - \frac{d}{dx} (y^2) + \frac{d}{dx} (x) = \frac{d}{dx} (2)$

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

For $x = 1 $ and $y = 2$, we obtain


$
\begin{equation}
\begin{aligned}
m &= \frac{-2(1)-2(2)-1}{2(1)-2(2)}\\
\\
m &= \frac{-2-4-1}{2-4}\\
\\
m &= \frac{-7}{-2}\\
\\
m &= \frac{7}{2}

\end{aligned}
\end{equation}
$


Using point slope form

$
\begin{equation}
\begin{aligned}
y-y_1 &= m(x-x_1)\\
\\
y-2 &= \frac{7}{2} (x-1)\\
\\
y- 2 &= \frac{7x-7}{2}\\
\\
y &= \frac{7x-7}{2}+2\\
\\
y &= \frac{7x-7+4}{2}\\
\\
y &= \frac{7x-3}{2} && \text{Equation of the tangent line at } (1,2)
\end{aligned}
\end{equation}
$