# Find dy/dx by implicit differentiation? √5x + y = 3 + x^2y^2

Apr 28, 2017

$y ' = \frac{2 x {y}^{2} - \sqrt{5}}{1 - 2 {x}^{2} y}$

or $\text{ } y ' = - \frac{2 x {y}^{2} - \sqrt{5}}{2 {x}^{2} y - 1}$

#### Explanation:

Given: $\sqrt{5} x + y = 3 + {x}^{2} {y}^{2}$

Use the product rule $\left(u v\right) ' = u v ' + v u '$

Let $u = {x}^{2} , u ' = 2 x \text{ and }$ Let $v = {y}^{2} , v ' = 2 y y '$

$\sqrt{5} + y ' = 0 + {x}^{2} \left(2 y y '\right) + {y}^{2} \left(2 x\right)$

$\sqrt{5} + y ' = 2 {x}^{2} y y ' + 2 x {y}^{2}$

Get the $y '$s on the same side of the equation and everything else on the other side of the equation:

$y ' - 2 {x}^{2} y y ' = 2 x {y}^{2} - \sqrt{5}$

Factor the $y '$:

$y ' \left(1 - 2 {x}^{2} y\right) = 2 x {y}^{2} - \sqrt{5}$

Isolate $y '$ by dividing:

$y ' = \frac{2 x {y}^{2} - \sqrt{5}}{1 - 2 {x}^{2} y}$

or $\text{ } y ' = - \frac{2 x {y}^{2} - \sqrt{5}}{2 {x}^{2} y - 1}$