Second Derivative Of A Parametric Function If $ y = 2t^3 + t^2 + 3$  $ x = t^2 + 2t + 1 $  then what is $d^2y \over dx^2$ for t = 1?
This is the question. What I tried is that, I first took the derivative using the rule $dy/dt \over dx/dt$, and then took the second derivative by taking the derivative of the result with respect to t. To formulate: $$\frac{d}{dt}(\frac{dy/dt}{dx/dt})$$
But in the end, the result didn't come out right. Is what I'm doing wrong? How to solve this question. 
Thanks in advance.
The answer is $5\over8$ by the way, so you'll know when you get it right.
 A: The formula for the second derivative of a parametric function is $$ \frac {\frac {d}{dt} (\frac {\frac {dy}{dt}}{\frac {dx}{dt}})} {\frac {dx}{dt}} $$.
Given this, we find that $\frac {dy}{dt} = 6t^2+2t$ and $\frac {dx}{dt}= 2t+2$. Thus, $\frac {dy}{dx} =\frac {3t^2+t}{t+1}$. Differentiating this with respect to $t$ yields $${\frac {d}{dt} (\frac {\frac {dy}{dt}}{\frac {dx}{dt}})}= \frac {3t^2+6t+1}{(t+1)^2}$$ 
Now, divide by $\frac {dx}{dt}$ to find that $\frac {d^2y}{dx^2}= \frac {3t^2+6t+1}{(t+1)^2(2t+2)}$.
Evaluating at $t=1$ yields $\frac {d^2y}{dx^2}|_{t=1} =\frac {3+6+1}{(4)(4)}= \frac{10}{16}= \frac {5}{8}$. 
A: You can do this on your own and you have proceeded in the right direction too. Just a small mistake in the last step. That is,
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$ since both $x$ and $y$ are functions of $t$ and then, 
$$\frac{d^2y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{dt}{dx}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\cdot \frac{1}{\frac{dx}{dt}}$$
Make that small correction and then whatever be the answer, your procedure is mathematically correct now and logical too.
