Question

How do you differentiate `-2y=x-cosy/(xy)`?

Answer 1

`frac{-x^2y^2 - y cos y}{x cos y + 2x^2y^2 + xy sin y} = y'(x)`

Explanation:

`-2 y' = 1 - frac{D_x cos y (xy) - cos y D_x (xy)}{x^2y^2}`

`-2 y' = 1 - frac{- y' sin y (xy) - cos y (y + xy')}{x^2y^2}`

`[-2x^2y^2] y' = x^2y^2 + y' [xy sin y] + y cos y + [x cos y]y'`

`-x^2y^2 - y cos y = [x cos y + 2x^2y^2 + xy sin y]y'`