Question

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

Answer 1

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+3xy`.

Note that the term `3xy` 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

`d/dx(y^3) = d/dx(x^3) + d/dx(3xy)`.

So `3y^2 dy/dx = 3x^2 + [3y+3x dy/dx]`

Solving for `dy/dx`, we have

`3y^2 dy/dx - 3x dy/dx = 3x^2 +3y`

`y^2 dy/dx - x dy/dx = x^2 +y`

`(y^2 - x) dy/dx = x^2 +y`

So,

`dy/dx = (x^2 +y)/(y^2 - x)`

If you are differentiating with respect to `t`

we get:

`3y^2 dy/dt = 3x^2 dx/dt +3y dx/dt +3x dy/dt`.

So,

`(y^2-x) dy/dx = (x^2+y) dx/dt`.