Question

What is `(d^2y)/(dy^2)`? `y^3+3x^4y-x^2=siny`

Answer 1

`(d^2y)/(dx^2) = ( (3y^2 + 3x^4 -cosy)(2-((12x^2(2x-12x^3y))/(3y^2 + 3x^4 -cosy))-24xy ) - (2x-6xy)(((6y(2x-12x^3y))/(3y^2 + 3x^4 -cosy)) + 12x^3 +siny((2x-12x^3y)/(3y^2 + 3x^4 -cosy))) ) / (3y^2 + 3x^2 -cosy)^2`

Explanation:

We have an implicit equation for `y`:

`y^3 + 3x^4y - x^2 = sin y`

We can differentiate wrt `x`:

`d/dx y^3 + d/dx3x^4y - d/dxx^2 = d/dxsin y`

For the second term we can use the product rule, and we can also explicitly apply the chain rule to differentiate any function of `y` wrt `x` (With practice we can differentiate implicitly, and omit these steps by differentiating wrt `y` and multiplying by `dy/dx`):

`dy/dxd/dyy^3 + (3x^4)(d/dx y) + (d/dx 3x^4)(y) - d/dxx^2 = dy/dxd/dysin y`

`:. dy/dx 3y^2 + 3x^4 dy/dx + 12x^3y - 2x = dy/dx cos y`

`:. dy/dx (3y^2 + 3x^4 -cosy) = 2x-12x^3y`

`:. dy/dx = (2x-12x^3y)/(3y^2 + 3x^4 -cosy)`

Now we can differentiate again wrt `x` (using the quotient rule) to get the second derivative, here we will use implicit differentiation:

`(d^2y)/(dx^2) = ( (3y^2 + 3x^4 -cosy)(d/dx(2x-12x^2y) ) - (2x-12x^3y)(d/dx (3y^2 + 3x^4 -cosy)) ) / (3y^2 + 3x^4 -cosy)^2`

`\ \ \ \ \ \ = ( (3y^2 + 3x^4 -cosy)(2-12x^2dy/dx-24xy ) - (2x-6xy)(6ydy/dx + 12x^3 +sinydy/dx) ) / (3y^2 + 3x^2 -cosy)^2`

The job of simplifying this expression is now algebra alone and by substituting our earlier result for `dy/dx` we can get a simplified expression

`(d^2y)/(dx^2) = ( (3y^2 + 3x^4 -cosy)(2-12x^2((2x-12x^3y)/(3y^2 + 3x^4 -cosy))-24xy ) - (2x-6xy)(6y((2x-12x^3y)/(3y^2 + 3x^4 -cosy)) + 12x^3 +siny((2x-12x^3y)/(3y^2 + 3x^4 -cosy))) ) / (3y^2 + 3x^2 -cosy)^2`

`\ \ \ \ \ \ = ( (3y^2 + 3x^4 -cosy)(2-((12x^2(2x-12x^3y))/(3y^2 + 3x^4 -cosy))-24xy ) - (2x-6xy)(((6y(2x-12x^3y))/(3y^2 + 3x^4 -cosy)) + 12x^3 +siny((2x-12x^3y)/(3y^2 + 3x^4 -cosy))) ) / (3y^2 + 3x^2 -cosy)^2`