Prove that $ y''(0) = -1, x = \cos\left(\frac{t}{1+t}\right), y = \sin\left(\frac{t}{1+t}\right)$ My attempt:
$$x'= -\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$
$$y'= \cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$
then I attempted to divide y' on x' which resulted in
$$-\cot\left(\frac{t}{1+t}\right)$$
When I try to take the double dash of this it gives me 
$$\csc^2\left(\frac{t}{1+t}\right),$$
but this function is undefined at 0.
 A: The second derivative of a parametric curves is
\begin{align*} 
&\frac{d}{dt} \left( \frac{dy}{dx} \right) / \left( \frac{dx}{dt} \right)\\
=&\frac{d}{dt} \left( -\cot \left( \frac{t}{1+t} \right) \right) / \left( \frac{d}{dt} \left( \sin\left( \frac{t}{1+t} \right) \right) \right)\\
=&\frac{\csc^2\left( \frac{t}{1+t} \right)}{\left( 1+t \right)^2} \cdot \frac{\left( 1+t \right)^2}{-\sin\left( \frac{t}{1+t} \right)}\\
=&-\csc^3\left( \frac{t}{1+t} \right)
\end{align*}
This is not defined for $t=0$. 
You can see this easily by $\frac{t}{1+t}\mapsto t$. Now you can more clearly see that it traces the unit circle with $t=0$ giving the point $(0,1)$. Intuitively, the slope is infinitely great at this point.
A: Assuming primes are with respect to t, direct differentiation of $ y'$ gives:
$-\dfrac {(2 (1 + t) \cos[t/(1 + t)] + \sin[t/(1 + t)])} {(1 + t)^4} $
which gives you a different result:
$ y^{''}(0)=-2  $
EDIT1:

For t > 0, the unit circle exists only on the arc $ [(0.6,0.8)-(1,0)] $, approximately given on the plot. Negative arguments of $t$  unexpectedly deliver a unit circular billiards board :)  ; the above plots were made on domain $t$: $[( 0 <t <12),( -2 <t <0 ) ]$. The trouble spot $ t=-1 $ is to be tackled with a change of variable.  
If I were you, I would offer a bounty for correct domain definitions to go out of the puzzling situation, so that $ y''(0) $ can be defined. At this moment  it appears that it cannot be defined (by the participants so far).
A: $$
\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$$
A: I think I have found the solution
$$x'= -\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$
$$y'= \cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}$$
$$y''= \frac{\frac{dy}{dx}(\cos\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2})}{-\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}}$$
$$y''= \frac{-\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}*\frac{1}{(1+t)^2}+-2(1+t)(\cos{\frac{t}{1+t}}*\frac{1}{(1+t)^2})}{-\sin\left(\frac{t}{1+t}\right) * \frac{1}{(1+t)^2}}$$
When I calculate this as it is it still has 0 on the bottom but when I cancel out the -sin up and down it gives me -1, is my solution correct?
Edit: I can't cancel it out because of the + I overlooked it, I still don't know the solution
Edit: I did another mistake, I should differentiate of dy/dx(dy/dt/dx/dt) / dx/dt not just dy/dt
A: I emailed my professor and he said that the question is incorrect at it is in fact undefined.  I am very sorry
A: To prove y''(0)=-1
Take the derivative of the second equation using the product, quotient and chain rules. Then replace $t$ with $0$ and evaluate.
For the other two you can integrate:
$\int -\sin(\frac{t}{1+t})*\frac{1}{(1+t)^2}*dt$
$u=\frac{t}{1+t}$
$du=\frac{1}{(1+t)^2}*dt$
$\int -\sin(u)*du$
Which I think you can evaluate :)
