How do you differentiate #-2y=y^2/(xsin(x-y)#?

Question

1458 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-19T14:44:45

#dy/dx=-(2sin(x-y)+2xcos(x-y))/(1-2xcos(x-y))#

We can rearrange and simplify to get:

#-2xsin(x-y)=y#

#d/dx[y]=d/dx[-2xsin(x-y)]#

#d/dx[y]=d/dx[-2x]sin(x-y)-2xd/dx[sin(x-y)]#

#d/dx[y]=-2sin(x-y)-2xd/dx[sin(x-y)]#

#d/dx[y]=-2sin(x-y)-2xcos(x-y)d/dx[x-y]#

#d/dx[y]=-2sin(x-y)-2xcos(x-y)(d/dx[x]-d/dx[y])#

Using the chqain rule we get that #d/dx=dy/dx*d/dy#

#dy/dxd/dy[y]=-2sin(x-y)-2xcos(x-y)(1-dy/dxd/dy[y])#

#dy/dx=-2sin(x-y)-2xcos(x-y)(1-dy/dx)#

#dy/dx=-2sin(x-y)-2xcos(x-y)+2xcos(x-y)dy/dx#

#dy/dx-2xcos(x-y)dy/dx=-2sin(x-y)-2xcos(x-y)#

#dy/dx[1-2xcos(x-y)]=-2sin(x-y)-2xcos(x-y)#

#dy/dx=-(2sin(x-y)+2xcos(x-y))/(1-2xcos(x-y))#