# How do you differentiate -1=xy^3-x^2y?

Feb 5, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y - {y}^{3}}{3 x {y}^{2} - {x}^{2}}$

#### Explanation:

When doing implicit differentiation, if differentiating $y$ with respect to $x$ then $x$ can be differentiated as usual (i.e. $\frac{d}{\mathrm{dx}} \left(x\right) = 1$).

But $y$ is a bit different. The derivative of $y$ alone will not be $1$ but $\frac{\mathrm{dy}}{\mathrm{dx}}$.

For any power or function of y the chain rule will have to be implemented. For example, the derivative of ${y}^{3}$ will be $3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$. To differentiate $- 1 = x {y}^{3} - {x}^{2} y$ simply differentiate term by term. So:

$\frac{d}{\mathrm{dx}} \left(- 1\right) = 0$
$\frac{d}{\mathrm{dx}} \left(x {y}^{3}\right) = {y}^{3} + 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$ (Don't forget the product rule!)
$\frac{d}{\mathrm{dx}} \left(- {x}^{2} y\right) = - 2 x y - {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

So putting these back into the given function; differentiating $- 1 = x {y}^{3} - {x}^{2} y$ will give us:

$0 = {y}^{3} + 3 x {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} - 2 x y - {x}^{2} \frac{\mathrm{dy}}{\mathrm{dx}}$

Rearrange this to get dy/dx on one side of the equation:

$0 = {y}^{3} - 2 x y + \left(3 x {y}^{2} - {x}^{2}\right) \frac{\mathrm{dy}}{\mathrm{dx}}$

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x y - {y}^{3}}{3 x {y}^{2} - {x}^{2}}$

And, thus our final solution.