Question

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

Answer 1

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

Explanation:

`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)]]`