What is the implicit derivative of 5=-yx^2-xy/(y-1)+y^2x5=yx2xyy1+y2x?

1 Answer
Jan 20, 2018

(dy)/(dx)=(2xy+y/(y-1)-y^2)/(-x^2-x/(y-1)^2+2yx)dydx=2xy+yy1y2x2x(y1)2+2yx

Explanation:

First we take the change in xx of both sides.

d/dx[5]=d/dx[-yx^2]-d/dx[(xy)/(y-1)]+d/dx[y^2x]ddx[5]=ddx[yx2]ddx[xyy1]+ddx[y2x]

0=-yd/dx[x^2]+x^2d/dx[-y]-y/(y-1)d/dx[x]+xd/dx[y/(y-1)]+y^2d/dx[x]+xd/dx[y^2]0=yddx[x2]+x2ddx[y]yy1ddx[x]+xddx[yy1]+y2ddx[x]+xddx[y2]

0=-y(2x)+x^2d/dx[-y]-y/(y-1)(1)+xd/dx[y/(y-1)]+y^2(1)+xd/dx[y^2]0=y(2x)+x2ddx[y]yy1(1)+xddx[yy1]+y2(1)+xddx[y2]

0=-2xy+x^2d/dx[-y]-y/(y-1)+xd/dx[y/(y-1)]+y^2+xd/dx[y^2]0=2xy+x2ddx[y]yy1+xddx[yy1]+y2+xddx[y2]

The chain rule tells us that d/dx=d/dyxx(dy)/(dx)ddx=ddy×dydx. That gives us:
0=-2xy+x^2(dy)/(dx)d/dy[-y]-y/(y-1)+x(dy)/(dx)d/dy[y/(y-1)]+y^2+x(dy)/(dx)d/dy[y^2]0=2xy+x2dydxddy[y]yy1+xdydxddy[yy1]+y2+xdydxddy[y2]

0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x(((y-1)d/dy(y)-(yd/dy(y-1)))/(y-1)^2)+y^2+(dy)/(dx)(2yx)0=2xy+dydx(x2)yy1+dydxx(y1)ddy(y)(yddy(y1))(y1)2+y2+dydx(2yx)

0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x(((y-1)(1)-y(1))/(y-1)^2)+y^2+(dy)/(dx)(2yx)0=2xy+dydx(x2)yy1+dydxx((y1)(1)y(1)(y1)2)+y2+dydx(2yx)

0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x((y-1-y)/(y-1)^2)+y^2+(dy)/(dx)(2yx)0=2xy+dydx(x2)yy1+dydxx(y1y(y1)2)+y2+dydx(2yx)

0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x((-1)/(y-1)^2)+y^2+(dy)/(dx)(2yx)0=2xy+dydx(x2)yy1+dydxx(1(y1)2)+y2+dydx(2yx)

0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)(-x)/(y-1)^2+y^2+(dy)/(dx)(2yx)0=2xy+dydx(x2)yy1+dydxx(y1)2+y2+dydx(2yx)

(dy)/(dx)(-x^2)+(dy)/(dx)(-x)/(y-1)^2+(dy)/(dx)(2yx)=2xy+y/(y-1)-y^2dydx(x2)+dydxx(y1)2+dydx(2yx)=2xy+yy1y2

(dy)/(dx)(-x^2-x/(y-1)^2+2yx)=2xy+y/(y-1)-y^2dydx(x2x(y1)2+2yx)=2xy+yy1y2

(dy)/(dx)=(2xy+y/(y-1)-y^2)/(-x^2-x/(y-1)^2+2yx)dydx=2xy+yy1y2x2x(y1)2+2yx