# How do you find (dy)/(dx) given x^2y^3-3y=x^2?

Oct 31, 2016

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

#### Explanation:

${x}^{2} {y}^{3} - 3 y = {x}^{2}$

Differentiating wrt $x$ gives;
$\frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{3}\right) - \frac{d}{\mathrm{dx}} \left(3 y\right) = \frac{d}{\mathrm{dx}} \left({x}^{2}\right)$
$\therefore \frac{d}{\mathrm{dx}} \left({x}^{2} {y}^{3}\right) - 3 \frac{\mathrm{dy}}{\mathrm{dx}} = 2 x$

We can use the product rule:
$\frac{d}{\mathrm{dx}} \left(u v\right) = u \frac{\mathrm{dv}}{\mathrm{dx}} + v \frac{\mathrm{du}}{\mathrm{dx}}$
for the 1st term to give:

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

And we use the chain rule so that we differentiate the 1st term wert $y$ rather than $x$ to give;

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