How do you find (d^2y)/(dx^2)d2ydx2 for -4y^2+4=4x^24y2+4=4x2?

1 Answer
Oct 16, 2016

(d^2y)/dx^2= -1/y^3d2ydx2=1y3

Explanation:

Use Implicit Differentiation:

-8y(dy/dx) = 8x8y(dydx)=8x

dy/dx = (-x)/ydydx=xy

(d^2y)/dx^2 = d/dx(dy/dx)d2ydx2=ddx(dydx)

(d^2y)/dx^2= (d((-x)/y))/dxd2ydx2=d(xy)dx

(d^2y)/dx^2= {-y - -x(dy/dx)}/y^2d2ydx2=yx(dydx)y2

(d^2y)/dx^2= {(-y^2)/y - -x((-x)/y)}/y^2d2ydx2=y2yx(xy)y2

(d^2y)/dx^2= -{y^2/y + -x((-x)/y)}/y^2d2ydx2=y2y+x(xy)y2

(d^2y)/dx^2= -{y^2/y + x^2/y}/y^2d2ydx2=y2y+x2yy2

(d^2y)/dx^2= -{y^2 + x^2}/y^3d2ydx2=y2+x2y3

From the original equation, y^2 + x^2 = 1 y2+x2=1:

(d^2y)/dx^2= -1/y^3d2ydx2=1y3