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

Question

Christopher Agresta · Accepted Answer

# (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)) #

# 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)) #

Paisley Johnson · Answer

undefined To find $\frac{{d^2y}}{{dx^2}}$, first, express $y$ explicitly in terms of $x$, then differentiate twice with respect to $x$, using the chain rule and product rule where necessary.

Given $y^2 = \frac{{x - 1}}{{x + 1}}$, solve for $y$ to get $y = \pm \sqrt{\frac{{x - 1}}{{x + 1}}}$.

Taking the derivative of $y$ with respect to $x$ gives:

$\frac{{dy}}{{dx}} = \pm \frac{1}{2} \left(\frac{{x - 1}}{{x + 1}}\right)^{-\frac{1}{2}} \cdot \frac{{(x + 1) - (x - 1)}}{{(x + 1)^2}}$

Now, differentiate $\frac{{dy}}{{dx}}$ with respect to $x$ to find $\frac{{d^2y}}{{dx^2}}$.

This process involves using the quotient rule and the chain rule. After simplification, you should arrive at the final expression for $\frac{{d^2y}}{{dx^2}}$.