Question

How do you find `y''` by implicit differentiation of `x^3+y^3=1` ?

Answer 1

Differentiate one step at a time, then when you have a `(d^2y)/dx^2` term (i.e. y'') or can easily get one, rearrange the equation.

Implicit differentiation is remarkably similar to "regular" differentiation. We just need to treat any term with a y in it slightly differently.

First, we differentiate both sides of the equation:
`d/dx(x^3+y^3) = d/dx(1)`

By the addition rule:
`d/dx(x^3+y^3) = d/dx(x^3)+d/dx(y^3)`

We know that `d/dxx^3 = 3x^2` and `d/dx1=0`, so
`3x^2+d/dx(y^3)=0`

Now we use the chain rule to implicitly find `d/dx(y^3)`:
Let `u = y^3`
Then `(du)/dy = 3y^2`
and since `(du)/dy*dy/dx = (du)/dx`
we have `d/dx(y^3) = 3y^2*dy/dx`.

So we have `3x^2+3y^2*dy/dx = 0`, and we can rearrange our equation to get `dy/dx = (-3x^2)/(3y^2)`.

From here, the next step is to again differentiate both sides, this time using the quotient rule:
`(d^2y)/dx^2=d/dx(-(3x^2)/(3y^2)) = -d/dx((3x^2)/(3y^2))`
`=(3y^2*d/dx(3x^2)-3x^2*d/dx(3y^2))/(3y^2)^2`
`=-(6x3y^2 - 3x^2*d/dx(3y^2))/(9y^4)`

To find `d/dx(3y^2)`, we again implicitly differentiate to get `6y*dy/dx`

And so:
`(d^2y)/dx^2 = -(18xy^2 - 18x^2y*dy/dx)/(9y^4)`

Finally, recall that `dy/dx = (-3x^2)/(3y^2)`

So: `(d^2y)/dx^2 = -(18xy^2 - 18x^2y*(-3x^2)/(3y^2))/(9y^4)`.