Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 2

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: $4x^2+9y^2=36$
$\displaystyle \frac{d}{dx} (4x^2) + \frac{d}{dx} (9y^2) = \frac{d}{dx} ( 36 )$
$\displaystyle 4 \frac{d}{dx} (x^2) + 9 \frac{d}{dx}(y^2) = \frac{d}{dx}(36)$
$\displaystyle (4)(2x) + (9) (2y) \left( \frac{dy}{dx}\right) = 0 $

$
\begin{equation}
\begin{aligned}
8 x + (18y) \frac{dy}{dx} & = 0 \\
\\
(18 y) \frac{dy}{dx} & = -8x\\
\\
\frac{\cancel{18y} \frac{dy}{dx}}{\cancel{18y}} & = \frac{-8x}{18y}\\
\\
\frac{dy}{dx} &= \frac{-8x}{18y}\\
\\
\frac{dy}{dx} &= \frac{-4x}{9y} \qquad \text{ or } \qquad y' = \frac{-4x}{9y} && \text{(Equation 1)}
\end{aligned}
\end{equation}
$


b.) Solving for $y$

$
\begin{equation}
\begin{aligned}
9y^2 &= 36-4x^2\\
\\
\frac{\cancel{9}y^2}{\cancel{9}} &= \frac{36-4x^2}{9}\\
\\
y^2 &= \frac{36-4x^2}{9}\\
\\
y &= \pm \sqrt{\frac{36-4x^2}{9}}\\
\\
y &= \pm \frac{\sqrt{4(9-x^2)}}{\sqrt{9}}\\
\\
y &= \pm \frac{2 \sqrt{9-x^2}}{3} && \text{(Equation 2)}\\
\\
\frac{d}{dx} (y) & = \pm \frac{d}{dx} \left( \frac{2\sqrt{9-x^2}}{3} \right)\\
\\
\frac{dy}{dx} & = \pm \left( \frac{2}{3} \right) \frac{d}{dx} (9-x^2)^{\frac{1}{2}}\\
\\
\frac{dy}{dx} & = \pm \left( \frac{\cancel{2}}{3} \right) \left( \frac{1}{\cancel{2}}\right) (9-x^2)^{\frac{-1}{2}} \cdot \frac{d}{dx} (9-x^2)\\
\\
\frac{dy}{dx} & = \pm \frac{1}{3}(9-x^2)^{\frac{-1}{2}} (0-2x)\\
\\
\frac{dy}{dx} & = \pm \frac{1}{3} (9-x^2)^{\frac{-1}{2}} (-2x)\\
\\
\frac{dy}{dx} & = \pm \frac{2x}{3\sqrt{9-x^2}} \qquad \text{ or } \qquad y' = \pm \frac{2x}{3\sqrt{9-x^2}}
\end{aligned}
\end{equation}
$


c.) Substituting Equation 2 in Equation 1

$
\begin{equation}
\begin{aligned}
y' &= \frac{-4x}{9y} && \text{(Equation 1)}\\
\\
y' &= \pm \frac{2}{3} \sqrt{9-x^2} && \text{(Equation 2)}\\
\\
y' &= \frac{-4x}{ 9 \left( \pm \frac{2}{3} \sqrt{9-x^2}\right)}\\
\\
y' &= \frac{4x}{\pm \frac{18}{3} \sqrt{9-x^2}}\\
\\
y' &= \frac{4x}{\pm 6 \sqrt{9-x^2}}\\
\\
y' &= \frac{2x}{3 \sqrt{9-x^2}}
\end{aligned}
\end{equation}
$


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