Question #43877

1 Answer
Feb 25, 2018

#dy/dx=(cos(x+y)-cos(y))/(-xsin(y)-cos(x+y))#

Explanation:

Differentiate both sides with respect to #x#. This means that every time we differentiate a term containing #y#, we should end up with an instance of #dy/dx#.

#d/dx(xcos(y))=d/dx(sin(x+y))#

#cos(y)d/dx(x)+xd/dxcos(y)=cos(x+y)d/dx(x+y)#

#-xsin(y)dy/dx+cos(y)=(1+dy/dx)cos(x+y)#

Solve for #dy/dx:#

#-xsin(y)dy/dx+cos(y)=cos(x+y)+dy/dxcos(x+y)# (Multiply out on the right side)

#-xsin(y)dy/dx-dy/dxcos(x+y)=cos(x+y)-cos(y)# (Isolate all terms containing #dy/dx# on the left side, move all other terms to the right)

#dy/dx(-xsin(y)-cos(x+y))=cos(x+y)-cos(y)# (Factor out #dy/dx#)

#dy/dx=(cos(x+y)-cos(y))/(-xsin(y)-cos(x+y))#
(Divide both sides by #-xsin(y)-cos(x+y)#)