Question

How do you find `(d^2y)/(dx^2)` given `y^2=(x-1)/(x+1)`?

Answer 1

`(d^2y)/dx^2 = -(2x-1) / ((x+1)^3(x-1)) * 1/sqrt((x-1)/(x+1))`

Alternatively:

`(d^2y)/dx^2 = -(2x-1) / ((x+1)^4((x-1)/(x+1))^(3/2))`

Explanation:

`y^2 = (x-1)/(x+1)`

Differentiating the LHS implicitly and the RHS using the quotient rule wrt `x` we have:

`\ \ \ \ 2ydy/dx = ( (x+1)(d/dx(x-1)) - (x-1)(d/dx(x+1)) ) / (x+1)^2`
`:. 2ydy/dx = ( (x+1)(1) - (x-1)(1) ) / (x+1)^2`
`:. 2ydy/dx = ( x+1-x+1 ) / (x+1)^2`
`:. 2ydy/dx = 2 / (x+1)^2`
`:. ydy/dx = 1 / (x+1)^2`

`( ("NB If we wanted an explicit expression for "dy/dx" we have" ), (dy/dx = 1 / ((x+1)^2sqrt((x-1)/(x+1))) " as " y=sqrt((x-1)/(x+1)) ) )`

We have established:

`\ \ \ \ ydy/dx = 1 / (x+1)^2`
`:. ydy/dx = (x+1)^-2`

Differentiating the LHS of [1] Implicitly with the product rule, and the RHS using the chain rule we get:

`(y)(d/dx dy/dx) + (d/dxy)(dy/dx) = -2(x+1)^-3(1)`
`:. y(d^2y)/dx^2 + dy/dxdy/dx = -2/(x+1)^3`
`:. y(d^2y)/dx^2 + (dy/dx)^2 = -2/(x+1)^3`
`:. y(d^2y)/dx^2 + (1 / ((x+1)^2sqrt((x-1)/(x+1))))^2 = -2/(x+1)^3`
`:. y(d^2y)/dx^2 + 1 / ((x+1)^4((x-1)/(x+1))) = -2/(x+1)^3`

We can rearrange, and cancel a factor of `(x+1)`:
`-y(d^2y)/dx^2 = 1 / ((x+1)^3(x-1)) + 2/(x+1)^3`
`:. -y(d^2y)/dx^2 = (1+2(x-1)) / ((x+1)^3(x-1))`
`:. -y(d^2y)/dx^2 = (1+2x-2) / ((x+1)^3(x-1))`
`:. -y(d^2y)/dx^2 = (2x-1) / ((x+1)^3(x-1))`
`:. y(d^2y)/dx^2 = -(2x-1) / ((x+1)^3(x-1))`

And, as before to get an explicit expression we divide by `y` to get
`:. (d^2y)/dx^2 = -(2x-1) / ((x+1)^3(x-1)) * 1/sqrt((x-1)/(x+1))`

We can also write this as;

`(d^2y)/dx^2 = -(2x-1) / ((x+1)^4(x-1)/(x+1)) * 1/((x-1)/(x+1))^(1/2)`
`(d^2y)/dx^2 = -(2x-1) / ((x+1)^4((x-1)/(x+1))^(3/2))`