Question

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

Answer 1

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)