Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 35

Determine $y''$ of $x^3 + y^3 = 1$ by using implicit differentiation.

Solving for 1st Derivative


$
\begin{equation}
\begin{aligned}

\frac{d}{dx} (x^3) + \frac{d}{dx} (y^3) =& \frac{d}{dx} (1)
\\
\\
3x^2 + 3y^2 \frac{dy}{dx} =& 0
\\
\\
3y^2 \frac{dy}{dx} =& -3x^2
\\
\\
\frac{\displaystyle \cancel{3y^2} \frac{dy}{dx}}{\cancel{3y^2}} =& \frac{-\cancel{3}x^2}{\cancel{3}y^2}
\\
\\
\frac{dy}{dx} =& \frac{-x^2}{y^2}

\end{aligned}
\end{equation}
$


Solving for the 2nd Derivative


$
\begin{equation}
\begin{aligned}

\frac{d^2y}{dx^2} =& \frac{\displaystyle y^2 \frac{d}{dx} (-x^2) - (-x^2) \frac{d}{dx} (y^2)}{(y^2)^2}
\\
\\
\frac{d^2y}{dx^2} =& \frac{\displaystyle (y^2)(-2x) - (-x^2)(2y) \frac{dy}{dx}}{y^4}
\\
\\
\frac{d^2y}{dx^2} =& \frac{\displaystyle -2xy^2 + 2x^2y \frac{dy}{dx}}{y^4}
\qquad \qquad \text{We know that $\large \frac{dy}{dx} = \frac{-x^2}{y^2}$}
\\
\\
\frac{d^2y}{dx^2} =& \frac{\displaystyle -2xy^2 + (2x^2y) \left( \frac{-x^2}{y^2} \right)}{y^4}
\\
\\
\frac{d^2y}{dx^2} =& \frac{\displaystyle -2xy^2 + \left( \frac{-2x^4}{y} \right)}{y^4}
\\
\\
\frac{d^2y}{dx^2} =& \frac{\displaystyle \frac{-2xy^3 - 2x^4}{y}}{y^4}
\\
\\
\frac{d^2y}{dx^2} =&\frac{-2x(y^3 + x^3)}{(y)(y^4)}
\qquad \qquad \text{We know that $x^3 + y^3 = 1$}
\\
\\
\frac{d^2y}{dx^2} =& \frac{-2x(1)}{y^5}
\\
\\
\frac{d^2y}{dx^2} =& \frac{-2x}{y^5}


\end{aligned}
\end{equation}
$