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

1 Answer
Feb 19, 2018

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

Explanation:

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])

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))