Question

Question #36537

Answer 1

`(d^2y)/dx^2 = (3x^2(y^4-x^4))/y^7 = (-3x^2)/y^7`

Explanation:

`x^4-y^4=1`

Differentiate both sides of the equation with respect to `x`

`4x^3 - 4y^3 dy/dx = 0`

So `dy/dx = x^3/y^3`

Now differentiate again w.r.t. `x`.

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

Get rid of the common factor `y^2`

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

Replace `dy/dx = x^3/y^3`

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

Clear the fraction in the numerator by multiplying by `y^3/y^3`

`= (3x^2y^4-3x^6)/y^7`

Factor the numerator.

`= (3x^2(y^4-x^4))/y^7`

Recognize the opposite of the initial expression.

`x^4-y^4=1 iff y^4-x^4=-1`