# Find `(dy)/(dx)` given that `y - x= sin(xy)`

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Differentiate `y-x=sinxy` : Use implicit differentiation

`d/(dx)[y-x]=d/dx[sinxy]`

`(dy)/(dx)-1=cosxy(y+x(dy)/(dx))`

`(dy)/(dx)-1=ycos(xy)+x(dy)/(dx)cos(xy)`

`(dy)/(dx)[1-xcos(xy)]=ycos(xy)+1`

`(dy)/(dx)=(ycos(xy)+1)/(1-xcos(xy))`

It is given that `y - x = sin(x*y)`

The derivative `dy/dx` can be determined using implicit differentiation.

`dy/dx - 1 = cos(x*y)*(x*(dy/dx) + y)`

=> `dy/dx - 1 = cos(x*y)*x*(dy/dx) + cos(x*y)*y`

=> `dy/dx - cos(x*y)*x*(dy/dx) = cos(x*y)*y + 1`

=> `dy/dx = (cos(x*y)*y + 1)/(1 - cos(x*y)*x)`

**The required derivative **`dy/dx = (cos(x*y)*y + 1)/(1 - cos(x*y)*x)`