Question

How do you find `(d^2y)/(dx^2)` for `5=x^2-2y^2`?

Answer 1

Deleted, because it was incorrect

Answer 2

I get `-5/(4y^3)`

Explanation:

`2y^2=x^2-5`

`4y dy/dx = 2x`

`dy/dx = x/(2y)`

`d/dx(dy/dx) = d/dx(x/(2y))`

`= ((1)(2y)-x(2(dy/dx)))/(2y)^2`

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

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

`= (y-x(x/(2y)))/(2y^2) * (2y)/(2y)`

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

We started with `5=x^2-2y^2`, so we have

`2y^2-x^2=-5`, making the second derivative,

`(d^2y)/dx^2 = -5/(4y^3)`

Answer 3

`y'' = -5/4 1/y^3`

Explanation:

`2y(x)^2-x^2+5=0->d/(dx)(2y(x)^2-x^2+5)=4y y'-2x=0`

`d/(dx)(2y y'-x)=2((y')^2+y y'')-1=0` so

`y''=(1/2-(y')^2)/y` but `y'=1/2x/y` so

`y'' = (2y^2-x^2)/(4y^3) = -5/4 1/y^3`