Question

How do you implicitly differentiate `2= xy-ysin^2x-cos^2xy`?

Answer 1

`y'= -(y + ysin2x +ysin2xy)/(x-sin^2x-xsin2xy)`

Explanation:

Differentiate the equation with respect to `x`:

`0 = d/(dx) (xy-ysin^2x-cos^2xy)`

`0 = y +xy'+ y'sin^2x + 2ysinxcosx+2cosxysinxy(y+xy')`

`0 = y +xy'+ y'sin^2x + ysin2x+ysin2xy+xy'sin2xy`

`-y'(x-sin^2x-xsin2xy)= y + ysin2x +ysin2xy`

`y'= -(y + ysin2x +ysin2xy)/(x-sin^2x-xsin2xy)`