Find ${{{d^2}y} \over {d{x^2}}}$ when $x = {1 \over {t + 1}}$ and $y = {1 \over {t - 1}}$ The answer in the text book is:
${{{d^2}y} \over {d{x^2}}} = 4{\left( {{{t + 1} \over {t - 1}}} \right)^3}$
I've tried this:
$\eqalign{
  & x = {1 \over {t + 1}},y = {1 \over {t - 1}}  \cr 
  & x = {(t + 1)^{ - 1}},y = {(t - 1)^{ - 1}}  \cr 
  & {{dx} \over {dt}} =  - {(t + 1)^{ - 2}} = {{ - 1} \over {{{(t + 1)}^2}}}  \cr 
  & {{dy} \over {dt}} =  - {(t - 1)^{ - 2}} = {{ - 1} \over {{{(t - 1)}^2}}}  \cr 
  & {{dy} \over {dx}} = {{dy} \over {dt}} \times \left( {{1 \over {{{dx} \over {dt}}}}} \right)  \cr 
  & {{dy} \over {dx}} = {{ - 1} \over {{{(t - 1)}^2}}} \times {{{{(t + 1)}^2}} \over { - 1}} = {{{{(t + 1)}^2}} \over {{{(t - 1)}^2}}} \cr} $

My approach to ${{{d^2}y} \over {d{x^2}}}$ is:
$${{dy} \over {dx}} = {{{{(t + 1)}^2}} \over {{{(t - 1)}^2}}}$$
So using the quotient rule:
$\eqalign{
  & u = {(t + 1)^2}  \cr 
  & v = {(t - 1)^2}  \cr 
  & {{du} \over {dt}} = 2(t + 1)  \cr 
  & {{dv} \over {dt}} = 2(t - 1)  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {{2{{(t - 1)}^2}(t + 1) - 2{{(t + 1)}^2}(t - 1)} \over {{{(t - 1)}^4}}}  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {{2(t - 1)(t + 1)\left[ {(t - 1) - (t + 1)} \right]} \over {{{(t - 1)}^4}}}  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {{2(t - 1)(t + 1)( - 2)} \over {{{(t - 1)}^4}}}  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {{ - 4(t + 1)} \over {{{(t - 1)}^3}}} \cr} $

As the previous attempt yielded an incorrect answer I attempted to answer it this way:
$\eqalign{
  & {{dx} \over {dt}} =  - {(t + 1)^{ - 2}}  \cr 
  & {{dy} \over {dt}} =  - {(t - 1)^{ - 2}}  \cr 
  & {{{d^2}x} \over {d{t^2}}} = 2{(t + 1)^{ - 3}}  \cr 
  & {{{d^2}y} \over {d{t^2}}} = 2{(t - 1)^{ - 3}} \cr} $
$\eqalign{
  & {{{d^2}y} \over {d{x^2}}} = {{{d^2}y} \over {d{t^2}}} \times \left( {{1 \over {{{{d^2}x} \over {d{t^2}}}}}} \right)  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {2 \over {{{(t - 1)}^3}}} \times {{{{(t + 1)}^3}} \over 2}  \cr 
  & {{{d^2}y} \over {d{x^2}}} = {{{{(t + 1)}^3}} \over {{{(t - 1)}^3}}} \cr} $

I'd appreciate it greatly if someone could point out my misunderstanding(s) and show me the correct approach. 
Thank you very much.
 A: You have computed $\frac{dy}{dx}$. That's a good start. 
In your first attempted calculation of $\frac{d^2y}{dx^2}$, you actually calculated $\frac{d}{dt}\left(\frac{dy}{dx}\right)$. That's good, but we want the derivative of $\frac{dy}{dx}$ with respect to $x$, not with respect to $t$.
Easily fixed! Multiply by $\frac{dt}{dx}$, or equivalently divide by $\frac{dx}{dt}$. You used that "trick" already, just use it once more. All of the ingredients you need are present in your first attempt.
Remark: There is no easy fix I can think of for the second attempt. Even though it is true that $\frac{dz}{dw}$ is the reciprocal of $\frac{dw}{dz}$, it is not in general true that  $\frac{d^2 z}{dw^2}$ is the reciprocal of $\frac{d^2 w}{dz^2}$. 
Why can't you take the reciprocal? Because when you "flip," the $2$'s end up in the wrong places! By the way, just in case of misunderstanding, the "explanation" in the preceding sentence  is a joke. 
A: As André said, your first computation was almost right, but $$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{\mathrm{d}}{\mathrm{d} x}\frac{\mathrm{d} y}{\mathrm{d} x} = \frac{\mathrm{d} t}{\mathrm{d} x}\frac{\mathrm{d}}{\mathrm{d} t}\frac{\mathrm{d} y}{\mathrm{d} x} = \left(\frac{\mathrm{d} x}{\mathrm{d} t}\right)^{-1}\frac{\mathrm{d}}{\mathrm{d} t}\frac{\mathrm{d} t}{\mathrm{d} x}$$

I just thought your second attempt is worth talking about - the problem is $$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} \neq \frac{\frac{\mathrm{d}^2 y}{\mathrm{d} t^2}}{\frac{\mathrm{d}^2 x}{\mathrm{d} t^2}}$$
Let's quickly look at $y=t^2$, $x=t$. Then $$\frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{\mathrm{d}^2}{\mathrm{d} t^2} t^2 = 2$$ Everything goes wrong when you see that $$\frac{\mathrm{d}^2 x}{\mathrm{d} t^2} = 0$$
Obviously looking at the second derivative isn't enough!
A: As an alternative to the implicit approach, we could write $y$ in terms of $x,$ then translate back into terms of $t$ at the end. Take $t\neq\pm 1$. Since $$x=\frac1{t+1},$$ then $$t+1=\frac1x,$$ so $$t-1=\frac1x-2=\frac{1-2x}{x},$$ and so $$y=\frac1{t-1}=\frac{x}{1-2x}.$$ Now, $$\frac{dy}{dx}=\frac{(1-2x)-(-2)x}{(1-2x)^2}=\frac{1}{(1-2x)^2}=(1-2x)^{-2},$$ so $$\frac{d^2y}{dx^2}=-2\cdot(1-2x)^{-3}\cdot\frac{d}{dx}[1-2x]=\frac4{(1-2x)^3},$$ and since $y=\frac{x}{1-2x},$ then $$\frac{d^2y}{dx^2}=4\cdot\frac{y^3}{x^3}=4\left(\frac yx\right)^3.$$ Finally, recalling that $x=\frac1{t+1}$ and $y=\frac1{t-1},$ it follows that $$\frac{d^2y}{dx^2}=4\left(\frac{t+1}{t-1}\right)^3.$$
A: One easy way is to use this: If $x(t),y(t)$ are differentiable functions of $t$ and $x'(t)\neq 0$, then
$$ \frac{dy}{dx}=\frac{y'(t)}{x'(t)}. $$
For this problem $x(t)=\frac{1}{t+1},y(t)=\frac{1}{t-1}$, then
\begin{eqnarray*}
\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{-\frac{1}{(t-1)^2}}{-\frac{1}{(t+1)^2}}=\frac{(t+1)^2}{(t-1)^2}=\left(\frac{t+1}{t-1}\right)^2
\end{eqnarray*}
and hence
\begin{eqnarray*}
\frac{d^2y}{dx^2}=\frac{d}{dx}\frac{dy}{dx}=\frac{\frac{d}{dt}\frac{dy}{dx}}{\frac{dx}{dt}}=\frac{\frac{-4(t+1)}{(t-1)^3}}{-\frac{1}{(t+1)^2}}=\frac{(t+1)^3}{(t-1)^3}.
\end{eqnarray*}
