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

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

#### Explanation:

Given equation:

$- y = \setminus \cos \frac{y}{x} - {x}^{2} / y$

$y = {x}^{2} / y - \setminus \cos \frac{y}{x}$

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

$\setminus \frac{d}{\mathrm{dx}} y = \setminus \frac{d}{\mathrm{dx}} \left({x}^{2} / y\right) - \setminus \frac{d}{\mathrm{dx}} \left(\setminus \cos \frac{y}{x}\right)$

$y ' = \setminus \frac{y \setminus \frac{d}{\mathrm{dx}} {x}^{2} - {x}^{2} \setminus \frac{d}{\mathrm{dx}} y}{{y}^{2}} - \setminus \frac{x \setminus \frac{d}{\mathrm{dx}} \setminus \cos y - \setminus \cos y \setminus \frac{d}{\mathrm{dx}} x}{{x}^{2}}$

$y ' = \setminus \frac{2 x y - {x}^{2} y '}{{y}^{2}} - \setminus \frac{- x \setminus \sin y \setminus y ' - \setminus \cos y}{{x}^{2}}$

${x}^{2} {y}^{2} y ' = 2 {x}^{3} y - {x}^{4} y ' + x {y}^{2} \setminus \sin y \setminus y ' + {y}^{2} \setminus \cos y$

$\left({x}^{4} + {x}^{2} {y}^{2} - x {y}^{2} \setminus \sin y\right) y ' = 2 {x}^{3} y + {y}^{2} \setminus \cos y$

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