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

##### 1 Answer

$y ' = \frac{3 {x}^{2} {y}^{2} - 2 x {y}^{3} - 2 {y}^{4}}{3 {x}^{2} {y}^{2} + 8 x {y}^{3} - 2 {x}^{3} y - 1}$

#### Explanation:

From the given $- y = {x}^{3} {y}^{2} - {x}^{2} {y}^{3} - 2 x {y}^{4}$
differentiate both sides of the equation with respect to x

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

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

$- y ' = {x}^{3} \cdot 2 y y ' + 3 {x}^{2} {y}^{2} - 3 {x}^{2} {y}^{2} y ' - 2 x {y}^{3} - 8 x {y}^{3} y ' - 2 {y}^{4}$

Isolate all terms with y' in one side of the equation

$- y ' - 2 {x}^{3} y y ' + 3 {x}^{2} {y}^{2} y ' + 8 x {y}^{3} y ' = 3 {x}^{2} {y}^{2} - 2 x {y}^{3} - 2 {y}^{4}$

Factor out the common monomial factor $y '$

$\left(- 1 - 2 {x}^{3} y + 3 {x}^{2} {y}^{2} + 8 x {y}^{3}\right) y ' = 3 {x}^{2} {y}^{2} - 2 x {y}^{3} - 2 {y}^{4}$

Divide both sides of the equation by the coefficient of $y '$ which is $\left(- 1 - 2 {x}^{3} y + 3 {x}^{2} {y}^{2} + 8 x {y}^{3}\right)$

$\frac{\left(- 1 - 2 {x}^{3} y + 3 {x}^{2} {y}^{2} + 8 x {y}^{3}\right) y '}{\left(- 1 - 2 {x}^{3} y + 3 {x}^{2} {y}^{2} + 8 x {y}^{3}\right)} = \frac{3 {x}^{2} {y}^{2} - 2 x {y}^{3} - 2 {y}^{4}}{- 1 - 2 {x}^{3} y + 3 {x}^{2} {y}^{2} + 8 x {y}^{3}}$

$y ' = \frac{3 {x}^{2} {y}^{2} - 2 x {y}^{3} - 2 {y}^{4}}{3 {x}^{2} {y}^{2} + 8 x {y}^{3} - 2 {x}^{3} y - 1}$

God bless....I hope the explanation is useful.