How do you implicitly differentiate #-1=-y^2x-2xy-xye^x #?
1 Answer
Apr 23, 2016
#dy/dx=(y^2+2y+xye^x+ye^x)/(-2xy-2x-xe^x)#
Explanation:
Given -
#-y^2x-2xy-xye^x=-1#
#[-y^2(1)+x(-2)y(dy)/(dx)]-[2xdy/dx(1)+2y]-[xye^x+xe^xdy/dx(1)+ye^x(1)]=0#
#-y^2-2xydy/dx-2xdy/dx-2y-xye^x-xe^xdy/dx-ye^x=0#
#-2xydy/dx-2xdy/dx-xe^xdy/dx=y^2+2y+xye^x+ye^x#
#dy/dx[-2xy-2x-xe^x]=y^2+2y+xye^x+ye^x#
#dy/dx=(y^2+2y+xye^x+ye^x)/(-2xy-2x-xe^x)#
