Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 36

Determine $y''$ of $x^4 + y^4 = a^4$ by using implicit
differentiation.

Solving for 1st Derivative, where $a$ is a constant


$\begin{equation} \begin{aligned} \frac{d}{dx} (x^4) + \frac{d}{dx} (y^4) =& \frac{d}{dx} (a^4) \\ \\ 4x^3 + 4y^3 \frac{dy}{dx} =& 0 \\ \\ 4y^3 \frac{dy}{dx} =& -4x^3 \\ \\ \frac{\displaystyle \cancel{4y^3} \frac{dy}{dx}}{\cancel{4y^3}} =& \frac{-\cancel{4}x^3}{\cancel{4}y^3} \\ \\ \frac{dy}{dx} =& \frac{-x^3}{y^3} \end{aligned} \end{equation}$


Solving for 2nd Derivative


$\begin{equation} \begin{aligned} \frac{d^2y}{dx^2} =& \frac{\displaystyle (y^3) \frac{d}{dx} (-x^3) - (-x^3) \frac{d}{dx} (y^3)}{(y^3)^2} \\ \\ \frac{d^2y}{dx^2} =& \frac{\displaystyle (y^3) (-3x^2) - (-x^3)(3y^2) \frac{dy}{dx}}{y^6} \\ \\ \frac{d^2y}{dx^2} =& \frac{\displaystyle -3x^2 y^3 + 3x^3 y^2 \frac{dy}{dx}}{y^6} \qquad \qquad \text{We know that $\large \frac{dy}{dx} = \frac{-x^3}{y^3}$} \\ \\ \frac{d^2y}{dx^2} =& \frac{\displaystyle -3x^2 y^3 + (3x^3y^2)\left( \frac{-x^3}{y^3} \right)}{y^6} \\ \\ \frac{d^2y}{dx^2} =& \frac{\displaystyle -3x^2 y^3 - \left( \frac{3x^6}{y} \right) }{y^6} \\ \\ \frac{d^2y}{dx^2} =& \frac{-3x^2 y^4 - 3x^6}{(y)(y^6)} \\ \\ \frac{d^2y}{dx^2} =& \frac{-3x^2 (y^4 + x^4)}{y^7} \qquad \qquad \text{We know that $x^4 + y^4 = a^4$} \\ \\ \frac{d^2y}{dx^2} =& \frac{-3x^2 (a^4)}{y^7} \\ \\ \frac{d^2y}{dx^2} =& \frac{-3a^4x^2}{y^7} \text{ or } y'' = \frac{-3a^4 x^2}{y^7} \end{aligned} \end{equation}$