# How do you find dy/dx by implicit differentiation given y=1/(x+y)?

Sep 11, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {y}^{2} / \left({y}^{2} + 1\right) , \text{ or, equivalently, } \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {\left(x + y\right)}^{2}} .$

#### Explanation:

We rewrite the given eqn. $y = \frac{1}{x + y} , \text{ as, } y \left(x + y\right) = 1 ,$

$i . e . , x y + {y}^{2} = 1.$

$\therefore x y = 1 - {y}^{2} , \mathmr{and} , x = \frac{1 - {y}^{2}}{y} .$

$\therefore x = \frac{1}{y} - {y}^{2} / y = \frac{1}{y} - y .$

Diff.ing both sides, w.r.t. $y ,$ we have,

$\frac{\mathrm{dx}}{\mathrm{dy}} = \frac{d}{\mathrm{dy}} \left(\frac{1}{y} - y\right) = - \frac{1}{y} ^ 2 - 1 = - \frac{1 + {y}^{2}}{y} ^ 2.$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\frac{\mathrm{dx}}{\mathrm{dy}}} = - {y}^{2} / \left({y}^{2} + 1\right) .$

Equivalently, this can be written as,

dy/dx=-y^2/(y^2+1)=-y^2/{y^2(1+1/y^2),

$= - \frac{1}{1 + \frac{1}{y} ^ 2} ,$ &, since, $\frac{1}{y} = \left(x + y\right) ,$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{1}{1 + {\left(x + y\right)}^{2}} ,$

Enjoy Maths.!