Hello,
I need to find the second derivative of 2x^2 - 3y^2 = 4 using implicit differentiation.
I got 2x/3y as my first derivative but am confused on how to solve for the second derivative using implicit differentiation.
+118609 This is what I would do...
\(\frac{dy}{dx}=\frac{2x}{3y}\\ \frac{d^2y}{dx^2}=\frac{(3y)(2)-(3\frac{dy}{dx})(2x)}{(3y)^2}\\ \frac{d^2y}{dx^2}=\frac{(y)(2)-(\frac{dy}{dx})(2x)}{3(y)^2}\\ \frac{d^2y}{dx^2}=\frac{2y-(\frac{2x}{3y})(2x)}{3y^2}\\ \frac{d^2y}{dx^2}=\frac{2y}{3y^2}-\frac{4x^2}{9y^3}\\ \frac{d^2y}{dx^2}=\frac{6y^2-4x^2}{9y^3}\\ \)
OR
\(2x^2 - 3y^2 = 4\\ 4x-6y\frac{dy}{dx}=0\\ 4-6[y(\frac{d^2y}{dx^2})+(\frac{dy}{dx}\cdot \frac{dy}{dx})=0\\ 4-6[y(\frac{d^2y}{dx^2})+\frac{2x}{3y}\cdot \frac{2x}{3y})=0\\ 4-6[y(\frac{d^2y}{dx^2})+\frac{4x^2}{9y^2}]=0\\ 4-6y(\frac{d^2y}{dx^2})-\frac{6*4x^2}{9y^2}=0\\ 3y(\frac{d^2y}{dx^2})=2-\frac{4x^2}{3y^2}\\ 3y(\frac{d^2y}{dx^2})=\frac{6y^2-4x^2}{3y^2}\\ \frac{d^2y}{dx^2}=\frac{6y^2-4x^2}{9y^3}\\\)
LaTex
\frac{dy}{dx}=\frac{2x}{3y}\\
\frac{d^2y}{dx^2}=\frac{(3y)(2)-(3\frac{dy}{dx})(2x)}{(3y)^2}\\
\frac{d^2y}{dx^2}=\frac{(y)(2)-(\frac{dy}{dx})(2x)}{3(y)^2}\\
\frac{d^2y}{dx^2}=\frac{2y-(\frac{2x}{3y})(2x)}{3y^2}\\
\frac{d^2y}{dx^2}=\frac{2y}{3y^2}-\frac{4x^2}{9y^3}\\
\frac{d^2y}{dx^2}=\frac{6y^2-4x^2}{9y^3}\\
