Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 1

a.) Determine $y'$ by Implicit Differentiation.
b.) Find the equation explicitly for $y$ and differentiate to get $y'$ in terms of $x$.
c.) Check that your solutions to part (a) and (b) are consistent by substituting the expression for $y$ into your solution for part (a).

a.) Given: $xy + 2x + 3x^2 = 4$
$\displaystyle \left[ x \frac{d}{dx} (y) + y \frac{d}{dx} (x) \right] + 2 \frac{d}{dx}(x) + 3 \frac{d}{dx} (x^2) = \frac{d}{dx} (4)$
$\displaystyle (x) \left( \frac{dy}{dx} \right) + y + 2 (3)(2x) = 0$
$\displaystyle (x) \left( \frac{dy}{dx} \right)+ y + 2 + 6x = 0$

$
\begin{equation}
\begin{aligned}
(x) \left( \frac{dy}{dx} \right) &= -y-2-6x\\
\\
\frac{(\cancel{x})\left( \frac{dy}{dx} \right)}{\cancel{x}} &= \frac{-y-2-6x}{x}\\
\\
\frac{dy}{dx} &= \frac{-y-2-6x}{x} \qquad \text{ or } \qquad y' = \frac{-y-2-6x}{ x} && \text{(Equation 1)}
\end{aligned}
\end{equation}
$



b.) Solving for $y$
$xy+2x+3x^2=4$

$
\begin{equation}
\begin{aligned}
xy &= 4 - 2x - 3x^2\\
\\
\frac{\cancel{x}y}{\cancel{x}} & = \frac{4-2x-3x^2}{x}\\
\\
y &= \frac{4-2x-3x^2}{x} \qquad \text{ or } \qquad y = \frac{4}{x} - 2 - 3x && \text{(Equation 2)}\\
\\
\frac{d}{dx}(y) &= \frac{d}{dx} \left( \frac{4}{x}\right) - \frac{d}{dx} (2) - \frac{d}{dx} (3x)\\
\\
\frac{dy}{dx} &= \left[ \frac{x \frac{d}{dx} (4) - 4 \frac{d}{dx} (x) }{x^2}\right] - 0 -3\\
\\
\frac{dy}{dx} &= \frac{(x)(0)-(4)(1)}{x^2}-3\\
\\
\frac{dy}{dx} &= \frac{-4}{x^2} -3 \qquad \text{ or } \qquad y' = \frac{-4}{x^2} -3
\end{aligned}
\end{equation}
$


c.) Substituting Equation 2 in Equation 1

$
\begin{equation}
\begin{aligned}
y' &= \frac{-y-2-6x}{x} && \text{(Equation 1)}\\
\\
y &= \frac{4}{x} - 2 - 3x && \text{(Equation 2)}\\
\\
y' &= \frac{-\left(\frac{4}{x}-2-3x\right) - 2 - 6x}{x}\\
\\
y' &= \frac{\frac{-4}{x}-\cancel{2} -3x -\cancel{2} -6x}{x}\\
\\
y' &= \frac{\frac{-4}{x} - 3x }{x}\\
\\
y' &= \frac{-4-3x^2}{(x)(x)}\\
\\
y' &= \frac{-4-3x^2}{x^2}\\
\\
y' &= \frac{-4}{x^2} -\frac{3\cancel{x^2}}{\cancel{x^2}}\\
\\
y' &= \frac{-4}{x^2} - 3\\
\end{aligned}
\end{equation}
$


Results from part (a) and part (b) are equivalent