# Question #e1184

Jan 22, 2018

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{3 y \left(y + 2 {x}^{2}\right)}{2 x \left({x}^{2} + 3 y\right)}$

#### Explanation:

$2 {x}^{3} y + 3 x {y}^{2} = 5$

First we take $\frac{d}{\mathrm{dx}}$ of every term.

$\frac{d}{\mathrm{dx}} \left[2 {x}^{3} y\right] + \frac{d}{\mathrm{dx}} \left[3 x {y}^{2}\right] = \frac{d}{\mathrm{dx}} \left[5\right]$

Then we use the product rule, $\frac{d}{\mathrm{dx}} \left[f \left(x\right) g \left(y\right)\right] = f \left(x\right) \frac{d}{\mathrm{dx}} \left[g \left(y\right)\right] + g \left(y\right) \frac{d}{\mathrm{dx}} \left[f \left(x\right)\right]$

$y \frac{d}{\mathrm{dx}} \left[2 {x}^{3}\right] + 2 {x}^{3} \frac{d}{\mathrm{dx}} \left[y\right] + 3 x \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] + {y}^{2} \frac{d}{\mathrm{dx}} \left[3 x\right] = \frac{d}{\mathrm{dx}} \left[5\right]$

$y \left(6 {x}^{2}\right) + 2 {x}^{3} \frac{d}{\mathrm{dx}} \left[y\right] + 3 x \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] + {y}^{2} \left(3\right) = 0$

the chain rule tells us that $\frac{d}{\mathrm{dx}} = \frac{d}{\mathrm{dy}} \times \frac{\mathrm{dy}}{\mathrm{dx}}$

$y \left(6 {x}^{2}\right) + \frac{\mathrm{dy}}{\mathrm{dx}} 2 {x}^{3} \frac{d}{\mathrm{dy}} \left[y\right] + \frac{\mathrm{dy}}{\mathrm{dx}} 3 x \frac{d}{\mathrm{dy}} \left[{y}^{2}\right] + {y}^{2} \left(3\right) = 0$

$6 {x}^{2} y + \frac{\mathrm{dy}}{\mathrm{dx}} 2 {x}^{3} \left(1\right) + \frac{\mathrm{dy}}{\mathrm{dx}} 3 x \left(2 y\right) + 3 {y}^{2} = 0$

$6 {x}^{2} y + \frac{\mathrm{dy}}{\mathrm{dx}} 2 {x}^{3} + \frac{\mathrm{dy}}{\mathrm{dx}} 6 x y + 3 {y}^{2} = 0$

Now we rearrange:
$\frac{\mathrm{dy}}{\mathrm{dx}} 2 {x}^{3} + \frac{\mathrm{dy}}{\mathrm{dx}} 6 x y = - 3 {y}^{2} - 6 {x}^{2} y$

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(2 {x}^{3} + 6 x y\right) = - 3 {y}^{2} - 6 {x}^{2} y$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{- 3 {y}^{2} - 6 {x}^{2} y}{2 {x}^{3} + 6 x y}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{3 {y}^{2} + 6 {x}^{2} y}{2 {x}^{3} + 6 x y}$

$\textcolor{w h i t e}{\frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{3 y \left(y + 2 {x}^{2}\right)}{2 x \left({x}^{2} + 3 y\right)}$