Question

How do you find y'' by implicit differentiation for `4x^3 + 3y^3 = 6`?

Answer 1

Please see below.

Explanation:

Starting with: `4x^3+3y^3=6`

Differentiate both sides with respect to `x`:

`12x^2+9y^2dy/dx=0`

Solve for `dy/dx` `" "` (See Note 1 , below)

`dy/dx = (-12x^2)/(9y^2)`, so

`dy/dx = (-4x^2)/(3y^2)`

Differentiate again, using the quotient rule to get

`(d^2y)/dx^2 = ((-8x)(3y^2)-(-4x^2)(6y dy/dx))/(3y^2)^2`

`= (-24xy^2+24x^2ydy/dx)/(9y^4)`

I prefer to remove the common factor before proceeding:

`= -24xy((y-x dy/dx)/(9y^4))`

Now, replace `dy/dx`

`= -24xy((y-x (-4x^2)/(3y^2))/(9y^4))`

`= -24xy((y + (4x^3)/(3y^2))/(9y^4))`

Now, simplify the complex fraction using your chosen technique.

`= -24xy(((y + (4x^3)/(3y^2))(3y^2))/((9y^4)(3y^2)))`

`= -24xy((3y^3 + 4x^3)/(27y^6))`

I see that I can reduce the fraction, but before I do there's a step I can do to simplify a lot.

Way back at the start of the problem, we were told that

`4x^3+3y^3=6`

So the numerator of our fraction is `6`. `" "` (See Note 2 below.)

`= -24xy((6)/(27y^6))`

Now simplify the quotient:

`(d^2y)/dx^2 = -(16x)/(3y^5)`

Note 1
Although we could differentiate again immediately, I prefer not to.

If we differentiate without solving for `dy/dx` first, we will need to be careful to disti nguish `(dy/dx)^2` from `(d^2y)/dx^2`. We get

`24x +18ydy/dx dy/dx +9y^2 (d^2y)/dx^2 = 0`.

It works, but it's kind of a mess.

Note 2
This step is typical of certain kinds of implicit differentiation second derivative problems. If you remember to look for it, it can simplify the result considerably.