Question

How do you find `(d^2y)/(dx^2)` for `3x^2+y^2=2`?

Answer 1

`(d^2y)/(dx^2) =- 6/y^3`

Explanation:

When we differentiate `y`, we get `dy/dx` When we differentiate a non implicit function of `y` then using the chain rule we can differentiate wrt `x` and multiply by `dy/dx`. When this is done in situ it is known as implicit differentiation.

We have:

`3x^2 + y^2 = 2`

Differentiate wrt `x` and we get:

`\ \ \ 6x + 2ydy/dx = 0`
`:. 3x + ydy/dx = 0`

And now (as we want the second derivative) we differentiate wrt x again, this time we must also apply the product rule:

`3 + (y)((d^2y)/(dx^2)) + (dy/dx)(dy/dx) = 0`
`:. 3 + y(d^2y)/(dx^2) + (dy/dx)^2 = 0`

We can rearrange the 1st derivative equation:

`:. dy/dx = -(3x)/y`

And if we substitute this into the 2nd derivative equation we get:

`:. 3 + y(d^2y)/(dx^2) + (-(3x)/y)^2 = 0`
`:. 3 + y(d^2y)/(dx^2) + (9x^2)/y^2 = 0`
`:. y(d^2y)/(dx^2) =- (9x^2)/y^2 -3`
`:. (d^2y)/(dx^2) =- (9x^2)/y^3 -3/y`
`" "=- (3(3x^2+y^2))/y^3`
`" "=- (3(2))/y^3`
`" "=- 6/y^3`

Answer 2

`(d^2y)/dx^2 = +- (-6)/(2-3x^2)^(3/2)`

Explanation:

We can solve the equation to make `y(x)` explicit:

`y(x) = +- sqrt(2-3x^2)`

So:

`dy/dx = +-(-3x)/(sqrt(2-3x^2))`

`(d^2y)/dx^2 = +- (-3sqrt(2-3x^2) -((-3x)(-6x)/(2sqrt(2-3x^2))))/(2-3x^2)`

`(d^2y)/dx^2 = +- (-3(2-3x^2)-9x^2 )/((2-3x^2)sqrt(2-3x^2))`

`(d^2y)/dx^2 = +- (-6+9x^2-9x^2 )/((2-3x^2)sqrt(2-3x^2))`

`(d^2y)/dx^2 = +- (-6)/((2-3x^2)sqrt(2-3x^2))`

We can also proceed implicitly by differentiating both sides of the equation with respect to `x`:

`d/dx (3x^2 + y^2) = d/dx (2)`

`6x +2ydy/dx = 0`

Solve for `dy/dx`:

`dy/dx = -(3x)/y`

Differentiate again:

`(d^2y)/dx^2 = (-3y +3xdy/dx)/y^2`

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

substituting `dy/dx` from the equation above:

`(d^2y)/dx^2 = (-9x^2)/y^3-3/y`

`(d^2y)/dx^2 = -(9x^2+3y^2)/y^3`

and as `y^2 = 2-3x^2`:

`(d^2y)/dx^2 = -(9x^2 +6 -9x^2)/y^3`

simplifying:

`(d^2y)/dx^2 = -6/y^3`

and substituting `y` from the original equation we get the same result.