Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 7

Determine $\displaystyle \frac{dy}{dx}$ of $x^2+xy-y^2=4$ by Implicit Differentiation.

$\displaystyle \frac{d}{dx} (x^2) + \frac{d}{dx} (xy) - \frac{d}{dx} (y^2) = \frac{d}{dx} (4)$


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


$
\begin{equation}
\begin{aligned}
xy' - 2yy' &= -2x -y\\
\\
y'(x-2y) &= -2x -y\\
\\
\frac{y'\cancel{(x-2y)}}{\cancel{(x-2y)}} &= \frac{-2x-y}{x-2y}\\
\\
y' &= \frac{-2x-y}{x-2y}

\end{aligned}
\end{equation}
$

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