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

Jan 21, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} {y}^{2} - 6 x {y}^{2} - 14 x {y}^{4}}{28 {x}^{2} {y}^{3} + 6 {x}^{2} y - 2 {x}^{3} y - 1}$

#### Explanation:

This implicit differentiation makes heavy use of the product rule. Before differentiating the function, we should do a sample first:

$\frac{d}{\mathrm{dx}} \left[5 {x}^{4} {y}^{2}\right] = 5 {y}^{2} \frac{d}{\mathrm{dx}} \left[{x}^{4}\right] + 5 {x}^{4} \frac{d}{\mathrm{dx}} \left[{y}^{2}\right] = 5 {y}^{2} \left(4 {x}^{3}\right) + 5 {x}^{4} \left(2 y\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

The most important part about this is that differentiating any term with a $y$ will spit out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ term, thanks to the chain rule.

Differentiating the given function:

$- \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} {y}^{2} + 2 {x}^{3} y \frac{\mathrm{dy}}{\mathrm{dx}} - 6 x {y}^{2} - 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 14 x {y}^{4} - 28 {x}^{2} {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}}$

Solve for $\frac{\mathrm{dy}}{\mathrm{dx}}$.

$28 {x}^{2} {y}^{3} \frac{\mathrm{dy}}{\mathrm{dx}} + 6 {x}^{2} y \frac{\mathrm{dy}}{\mathrm{dx}} - 2 {x}^{3} y \frac{\mathrm{dy}}{\mathrm{dx}} - \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} {y}^{2} - 6 x {y}^{2} - 14 x {y}^{4}$

Factor out a $\frac{\mathrm{dy}}{\mathrm{dx}}$ and divide:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} {y}^{2} - 6 x {y}^{2} - 14 x {y}^{4}}{28 {x}^{2} {y}^{3} + 6 {x}^{2} y - 2 {x}^{3} y - 1}$