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

1 Answer
Jan 20, 2018

#(dy)/(dx)=(2xy+y/(y-1)-y^2)/(-x^2-x/(y-1)^2+2yx)#

Explanation:

First we take the change in #x# of both sides.

#d/dx[5]=d/dx[-yx^2]-d/dx[(xy)/(y-1)]+d/dx[y^2x]#

#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=-y(2x)+x^2d/dx[-y]-y/(y-1)(1)+xd/dx[y/(y-1)]+y^2(1)+xd/dx[y^2]#

#0=-2xy+x^2d/dx[-y]-y/(y-1)+xd/dx[y/(y-1)]+y^2+xd/dx[y^2]#

The chain rule tells us that #d/dx=d/dyxx(dy)/(dx)#. 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+(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+(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+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x((y-1-y)/(y-1)^2)+y^2+(dy)/(dx)(2yx)#

#0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)x((-1)/(y-1)^2)+y^2+(dy)/(dx)(2yx)#

#0=-2xy+(dy)/(dx)(-x^2)-y/(y-1)+(dy)/(dx)(-x)/(y-1)^2+y^2+(dy)/(dx)(2yx)#

#(dy)/(dx)(-x^2)+(dy)/(dx)(-x)/(y-1)^2+(dy)/(dx)(2yx)=2xy+y/(y-1)-y^2#

#(dy)/(dx)(-x^2-x/(y-1)^2+2yx)=2xy+y/(y-1)-y^2#

#(dy)/(dx)=(2xy+y/(y-1)-y^2)/(-x^2-x/(y-1)^2+2yx)#