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

# 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)) #