Question

How do you differentiate `-y=cosy/x-x^2/y`?

Answer 1

`y'=\frac{2x^3y+y^2\cos y}{x^4+x^2y^2-xy^2\sin y}`

Explanation:

Given equation:

`-y=\cos y/x-x^2/y`

`y=x^2/y-\cos y/x`

differentiating the above equation w.r.t. `x` on both sides as follows

`\frac{d}{dx}y=\frac{d}{dx}(x^2/y)-\frac{d}{dx}(\cos y/x)`

`y'=\frac{y\frac{d}{dx}x^2-x^2\frac{d}{dx}y}{y^2}-\frac{x\frac{d}{dx}\cos y-\cos y\frac{d}{dx}x}{x^2}`

`y'=\frac{2xy-x^2y'}{y^2}-\frac{-x\siny\ y'-\cos y}{x^2}`

`x^2y^2y'=2x^3y-x^4y'+xy^2\sin y\ y'+y^2\cos y`

`(x^4+x^2y^2-xy^2\sin y)y'=2x^3y+y^2\cos y`

`y'=\frac{2x^3y+y^2\cos y}{x^4+x^2y^2-xy^2\sin y}`