If $$x^4+y^4=16$$​, then find the second derivative of $$y$$​.

Correct option is A

​$$\begin{aligned}&\text{Differentiate implicitly with respect to } x: \\&\quad 4x^3 + 4y^3 \frac{dy}{dx} = 0 \\[10pt]&\text{Solve for } \frac{dy}{dx}: \\&\quad \frac{dy}{dx} = -\frac{x^3}{y^3} \\[10pt]&\textbf{Second Derivative } \left(\frac{d^2y}{dx^2}\right) \\&\text{Differentiate } \frac{dy}{dx} = -\frac{x^3}{y^3} \text{ using the quotient rule:} \\&\quad \frac{d^2y}{dx^2} = \frac{-3x^2 y^3 - x^3 (3y^2 \frac{dy}{dx})}{y^6} \\[10pt]&\text{Substitute } \frac{dy}{dx} = -\frac{x^3}{y^3}: \\&\quad \frac{d^2y}{dx^2} = \frac{-3x^2 y^3 + 3x^6 / y^3}{y^6} \\[10pt]&\text{Combine terms and simplify using } x^4 + y^4 = 16: \\&\quad \frac{d^2y}{dx^2} = -\frac{48x^2}{y^7}\end{aligned}$$​​