Question

2nd derivative of x^2+4y^2=1

Answer 1

`\bf x^2+4y^2=1\\\\ -----------------------------\\\\ \textit{using implicit differentiation} \\\\\\ 2x+4\cdot 2y\cfrac{dy}{dx}=0\implies x+4y\cfrac{dy}{dx}=0\implies \boxed{\cfrac{dy}{dx}=\cfrac{-x}{4y}}\\\\ -----------------------------\\\\ \cfrac{dy^2}{dx^2}=\cfrac{-1\cdot 4y-(-x)4(dy)/(dx)}{(4y)^2}\impliedby \textit{using the quotient rule} \\\\\\ \cfrac{dy^2}{dx^2}=\cfrac{-4y+4x(dy)/(dx)}{4^2y^2}\implies \cfrac{dy^2}{dx^2}=\cfrac{-y+x(dy)/(dx)}{4y^2}\impliedby \textit{common factor}`


`\bf \cfrac{dy^2}{dx^2}=\cfrac{-y+x\cdot \boxed{(-x)/(4y)}}{4y^2}\implies \cfrac{dy^2}{dx^2}=\cfrac{-y-(x^2)/(4y)}{4y^2} \\\\\\ \cfrac{dy^2}{dx^2}=\cfrac{(-4y^2-x^2)/(4y)}{4y^2}\implies \cfrac{dy^2}{dx^2}=\cfrac{-4y^2-x^2}{4y}\cdot \cfrac{1}{4y^2} \\\\\\ \cfrac{dy^2}{dx^2}=\cfrac{-4y^2-x^2}{16y^3}\iff -\cfrac{1}{4y}-\cfrac{x^2}{16y^3}\impliedby \begin{array}{llll} \textit{distributing the}\\ \textit{denominator} \end{array}`