1. ## implicit differentiation

(1/xy)=x+y
find y'

2. $\frac{1}{xy} = x+y$ is the same as $xy(x+y) = 1$

$\frac{d}{dx} \left(xy(x+y)\right) = \frac{d}{dx} \left(1 \right)$

use the product rule

$xy \frac{d}{dx} \left(x+y \right) + (x+y)\frac{d}{dx} \left(xy \right) = 0$

Note: by the chain rule, $\frac{d}{dx} \left(f(y) \right) = f'(y)\frac{dy}{dx}$, so

$xy(1+\frac{dy}{dx})+(x+y)(y+x\frac{dy}{dx}) = 0$

$xy+xy\frac{dy}{dx}+xy+x^{2}\frac{dy}{dx}+y^{2}+xy\ frac{dy}{dx} = 0$

$(x^{2}+2xy)\frac{dy}{dx} = -y^{2}-2xy$

$\frac{dy}{dx} = -\frac{y^{2}+2xy}{x^2+2xy}$