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

##### 1 Answer
Jan 26, 2016

Please see the explanation section below.

#### Explanation:

Begin by rewriting. Add $y$ to both sides to get rid of the duplicate terms.

${y}^{3} = {x}^{3} + 3 x y$.

Note that the term $3 x y$ has a product of both variables in it. When we differentiate that term we will need the product rule.

Now, assuming that you are differentiating with respect to $x$, we get

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

So $3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dx}} = 3 {x}^{2} + \left[3 y + 3 x \frac{\mathrm{dy}}{\mathrm{dx}}\right]$

Solving for $\frac{\mathrm{dy}}{\mathrm{dx}}$, we have

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

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

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

So,

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

If you are differentiating with respect to $t$

we get:

$3 {y}^{2} \frac{\mathrm{dy}}{\mathrm{dt}} = 3 {x}^{2} \frac{\mathrm{dx}}{\mathrm{dt}} + 3 y \frac{\mathrm{dx}}{\mathrm{dt}} + 3 x \frac{\mathrm{dy}}{\mathrm{dt}}$.

So,

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