What is the implicit derivative of #1=xy-sinxy#?

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Answer 1 · 2016-03-20T17:00:14

#dy/dx=(y[1-cos(xy)]]/[[(cos(xy)-x)]]#

#1=xy-sin(xy)#

Differentiate both sides with respect to #xrArr#

#D(0)=D[xy-sin(xy)]#

#0=x.dy/dx+y-[cos(xy)xx(x.dy/dx+y)]#

#0=x.dy/dx+y-[cos(xy).xdy/dx+ycos(xy)]#

#0=x.dy/dx+y-cos(xy).dy/dx-ycos(xy)#

#cos(xy).dy/dx-x.dy/dx=y-ycos(xy)#

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

#:.dy/dx=(y[1-cos(xy)]]/[[(cos(xy)-x)]]#