# How do you implicitly differentiate -y= x^3y^2-3x^2y^2+xy^4 ?

##### 1 Answer
Nov 28, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x {y}^{2} - {y}^{4} - 3 {x}^{2} {y}^{2}}{1 + 2 {x}^{3} y - 6 {x}^{2} y + 4 x {y}^{3}}$

#### Explanation:

$- y = {x}^{3} {y}^{2} - 3 {x}^{2} {y}^{2} + x {y}^{4}$

We differentiate everything wrt $x$:

$- \frac{d}{\mathrm{dx}} \left(y\right) = \frac{d}{\mathrm{dx}} \left({x}^{3} {y}^{2}\right) - \frac{d}{\mathrm{dx}} \left(3 {x}^{2} {y}^{2}\right) + \frac{d}{\mathrm{dx}} \left(x {y}^{4}\right)$

We can just deal with the first term;

$- \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({x}^{3} {y}^{2}\right) - \frac{d}{\mathrm{dx}} \left(3 {x}^{2} {y}^{2}\right) + \frac{d}{\mathrm{dx}} \left(x {y}^{4}\right)$

For the other term we apply the product rule;

$- \frac{\mathrm{dy}}{\mathrm{dx}} = \left\{\left({x}^{3}\right) \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) + \left(\frac{d}{\mathrm{dx}} {x}^{3}\right) \left({y}^{2}\right)\right\} - \left\{\left(3 {x}^{2}\right) \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) + \left(\frac{d}{\mathrm{dx}} 3 {x}^{2}\right) \left({y}^{2}\right)\right\} + \left\{\left(x\right) \left(\frac{d}{\mathrm{dx}} {y}^{4}\right) + \left(\frac{d}{\mathrm{dx}} x\right) \left({y}^{4}\right)\right\}$

$\therefore - \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) + 3 {x}^{2} {y}^{2} - 3 {x}^{2} \left(\frac{d}{\mathrm{dx}} {y}^{2}\right) + 6 x {y}^{2} + x \left(\frac{d}{\mathrm{dx}} {y}^{4}\right) + {y}^{4}$

Next we use the chain rule so that we can differentiate wrt $y$ (this is the "Implicit" part of the differentiation
$\therefore - \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} \left(\frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} {y}^{2}\right) + 3 {x}^{2} {y}^{2} - 3 {x}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} {y}^{2}\right) - 6 x {y}^{2} + x \left(\frac{\mathrm{dy}}{\mathrm{dx}} \frac{d}{\mathrm{dy}} {y}^{4}\right) + {y}^{4}$

And now we can differentiate the $y$ terms wrt $y$
$\therefore - \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{3} \left(\frac{\mathrm{dy}}{\mathrm{dx}} 2 y\right) + 3 {x}^{2} {y}^{2} - 3 {x}^{2} \left(\frac{\mathrm{dy}}{\mathrm{dx}} 2 y\right) - 6 x {y}^{2} + x \left(\frac{\mathrm{dy}}{\mathrm{dx}} 4 {y}^{3}\right) + {y}^{4}$

Finally, we can tidy up, collect terms and factorise $\frac{\mathrm{dy}}{\mathrm{dx}}$ to form an explicit solution

$\therefore - \frac{\mathrm{dy}}{\mathrm{dx}} = 2 {x}^{3} y \frac{\mathrm{dy}}{\mathrm{dx}} + 3 {x}^{2} {y}^{2} - 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x {y}^{2} + 4 x {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + {y}^{4}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} + 2 {x}^{3} y \frac{\mathrm{dy}}{\mathrm{dx}} - 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} + 4 x {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} = 6 x {y}^{2} - {y}^{4} - 3 {x}^{2} {y}^{2}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left(1 + 2 {x}^{3} y - 6 {x}^{2} y + 4 x {y}^{3}\right) = 6 x {y}^{2} - {y}^{4} - 3 {x}^{2} {y}^{2}$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{6 x {y}^{2} - {y}^{4} - 3 {x}^{2} {y}^{2}}{1 + 2 {x}^{3} y - 6 {x}^{2} y + 4 x {y}^{3}}$