# How do you implicitly differentiate -y= x^3y^2-3x^2y^3-7xy^4 ?

Feb 17, 2016

A useful timesaver is to use $y '$ instead of $\frac{\mathrm{dy}}{\mathrm{dx}}$ when differentiating y ... here is how it works:

#### Explanation:

$- y = {x}^{3} {y}^{2} - 3 {x}^{2} {y}^{3} - 7 x {y}^{4}$

Now, using the chain and product rules ...

$- y ' = 3 {x}^{2} {y}^{2} + {x}^{3} \left(2 y y '\right) - 6 x {y}^{3} - 3 {x}^{2} \left(3 {y}^{2} y '\right) - 7 {y}^{4} - 7 x \left(4 {y}^{3} y '\right)$

Next combine the $y '$ terms on the left side of the equation ...

$y ' \left[- 1 - {x}^{3} \left(2 y\right) + 3 {x}^{2} \left(3 {y}^{2}\right) + 7 x \left(4 {y}^{3}\right)\right] = 3 {x}^{2} {y}^{2} - 6 x {y}^{3} - 7 {y}^{4}$

Finally, simplify and solve for $y '$

$y ' = \frac{3 {x}^{2} {y}^{2} - 6 x {y}^{3} - 7 {y}^{4}}{9 {x}^{2} \left({y}^{2}\right) + 28 x \left({y}^{3}\right) - {x}^{3} \left(2 y\right) - 1}$

hope that helped