Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 33

Determine $y''$ of $9x^2 + y^2 = 9$ by using implicit differentiation.

Solving for the 1st Derivative


$
\begin{equation}
\begin{aligned}

9 \frac{d}{dx} (x^2) + \frac{d}{dx} (y^2) =& \frac{d}{dx} (9)
\\
\\
(9)(2x) + (2y) \frac{dy}{dx} =& 0
\\
\\
18x + 2y \frac{dy}{dx} =& 0
\\
\\
2y \frac{dy}{dx} =& -18x
\\
\\
\frac{\displaystyle \cancel{2y} \frac{dy}{dx}}{\cancel{2y}} =& \frac{-18x}{2y}
\\
\\
\frac{dy}{dx} =& \frac{-9x}{y}

\end{aligned}
\end{equation}
$


Solving for the 2nd Derivative


$
\begin{equation}
\begin{aligned}

\frac{d^2y}{dx^2} =& \frac{\displaystyle y \frac{d}{dx} (-9x) - (-9x) \frac{d}{dx} (y)}{y^2}
&& \text{Apply Quotient Rule}
\\
\\
\frac{d^2y}{dx^2} =& \frac{\displaystyle (y)(-9) - (-9x) \frac{dy}{dx}}{y^2}
&& \text{Substitute $\large \frac{dy}{dx} = \frac{-9x}{y}$}
\\
\\
\frac{d^2y}{dx^2} =& \frac{-9y + (9x) \displaystyle \left( \frac{-9x}{y} \right) }{y^2}
&&
\\
\\
\frac{d^2y}{dx^2} =& \frac{-9y + \displaystyle \frac{(-81x^2)}{y}}{y^2}
&&
\\
\\
\frac{d^2y}{dx^2} =& \frac{\displaystyle \frac{-9y^2 - 81x^2}{y}}{y^2}
&&
\\
\\
\frac{d^2y}{dx^2} =& \frac{-9y^2 - 81x^2}{(y)(y^2)}
&&
\\
\\
\frac{d^2y}{dx^2} =& \frac{-9 (9x^2 + y^2)}{y^3}
&& \text{We know that $9x^2 + y^2 = 9$}
\\
\\
\frac{d^2y}{dx^2} =& \frac{-9(9)}{y^3}
&&
\\
\\
\frac{d^2y}{dx^2} =& \frac{-81}{y^3} \text{ or } y'' = \frac{-81}{y^3}

\end{aligned}
\end{equation}
$