Use this equation to find dy/dx. 4y cos (x) = x2 + y2

Let's differentiate implicitly,

don't worry about solving for y explicitly before differentiating.

(4y) cosx = x^2 + y^2

Apply product rule on the left,

and power rule on the right-hand terms,

(4y) 'cosx + 4y (cosx) ' = (x^2) ' + (y^2) '

4y'cosx + 4y (-sinx) = 2x + 2yy'

4y'cosx - 4ysinx = 2x + 2yy'

Let's get both y' terms on the left side,

and everything else on the right,

4y'cosx - 2yy' = 2x + 4ysinx

Factor y' out of each term,

y' (4cosx - 2y) = 2x + 4ysinx

Solve for y' (which is dy/dx) by dividing by the stuff,

y' = (2x + 4ysinx) / (4cosx - 2y)

You can divide a 2 out of everything to simplify it down a little bit.

y' = (x + 2ysinx) / (2cosx - y)