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.
This is what I would do...
dydx=2x3yd2ydx2=(3y)(2)−(3dydx)(2x)(3y)2d2ydx2=(y)(2)−(dydx)(2x)3(y)2d2ydx2=2y−(2x3y)(2x)3y2d2ydx2=2y3y2−4x29y3d2ydx2=6y2−4x29y3
OR
2x2−3y2=44x−6ydydx=04−6[y(d2ydx2)+(dydx⋅dydx)=04−6[y(d2ydx2)+2x3y⋅2x3y)=04−6[y(d2ydx2)+4x29y2]=04−6y(d2ydx2)−6∗4x29y2=03y(d2ydx2)=2−4x23y23y(d2ydx2)=6y2−4x23y2d2ydx2=6y2−4x29y3
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}\\