# Use implicit differentiationIf y=x+sin(xy) then dy/dx=

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You need to differentiate the function with respect to x such that:

`(dy)/(dx) = x' + (sin(xy))'*(xy)'`

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

`(dy)/(dx) = 1 + cos(xy)*(y + xy')`

You need to substitute `(dy)/(dx` ) for`y'` such that:

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

Opening the brackets yields:

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

You need to isolate to the left side terms containing `(dy)/(dx)` such that:

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

Factoring out `(dy)/(dx)` yields:

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

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

**Hence, evaluating `(dy)/(dx)` using implicit differentiation yields `(dy)/(dx) = (1 + y*cos(xy))/(1 - x*cos(xy)).` **