Use Implicit Differentiation to find $\frac{d^2y}{dx^2}$? Given a system of equation,
\begin{align*}
x &= t^2 + 2t \\
y &= 3t^4 + 4t^3
\end{align*}
I want to find $\frac{d^2 y}{dx^2}$ at $(x,y) = (8, 80)$. Then, $\partial_x(y) = \frac{d y}{dt} \frac{dt}{dx}$. By chain rule,
\begin{align*}
\partial_x^2(y) &= \partial_x \left(\frac{d y}{dt}\right)\frac{dt}{dx} + \frac{dy}{dt} \partial_x \left(\frac{dt}{dx}\right) \\
&= \frac{d^2 y}{dt^2}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\frac{d^2t}{dx^2}
\end{align*}
Here, how do I find $\frac{d^2 t}{dx^2}$?
 A: By implicit differentiation on $x = t^2 + 2t$, one obtains
$$1 = 2t \cdot \dfrac{\text{d}t}{\text{d}x} + 2\dfrac{\text{d}t}{\text{d}x} = \dfrac{\text{d}t}{\text{d}x}\left(2t+2 \right)$$
hence
$$\dfrac{\text{d}t}{\text{d}x} = \dfrac{1}{2t+2}\text{.}$$
Differentiating a second time on the equation $x = t^2 + 2t$, one obtains by an application of the product rule
$$0 = 2\left(t \cdot \dfrac{\text{d}^2t}{\text{d}x^2} + 1 \cdot \dfrac{\text{d}t}{\text{d}x}  \right) + 2\text{.}$$
It follows from the above that
$$\dfrac{\text{d}^2t}{\text{d}x^2} = \dfrac{\dfrac{-2}{2} - \dfrac{\text{d}t}{\text{d}x}}{t} = \dfrac{-1 - \dfrac{\text{d}t}{\text{d}x}}{t} = -\left(\dfrac{1 + \dfrac{\text{d}t}{\text{d}x}}{t} \right)\text{.}$$
Substitute $\dfrac{\text{d}t}{\text{d}x}$ as found earlier, and you're done.
Note that you will have to solve for $t$ to compute this quantity. (Hint: make use of the fact that $x = t^2 = 2t$ and that you have the point $(x, y) = (8, 80)$.)
A: $x=t^2+2t$
$y=3t^4+4t^3$
$\displaystyle\frac{dx}{dt}=2t+2$ and $\frac{dy}{dt}=12t^3+12t^2$
$\displaystyle\frac{dy}{dx}=\frac{12t^3+12t^2}{2t+2}=\frac{6t^3+6t^2}{t+1}=6t^2$
$\displaystyle\frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{dy}{dx})=\frac{d}{dx}6t^2=12t\frac{dt}{dx}=\frac{12t}{\frac{dx}{dt}}=\frac{12t}{2t+2}=\frac{6t}{t+1}$
Now when $x=8$, $t=-4,2$ and therefore corresponding values of $y=512,80$, Since we need to calculate $\frac{d^2y}{dx^2}$ at the point $(8,80)$ which corresponds to $t=2$ therefore ,
$\displaystyle\frac{d^2y}{dx^2}=\frac{6t}{t+1}=\frac{6*2}{2+1}=\fbox4$
A: 
The solution given by Deepak follows the usual way that the second derivative (parametric) function would be calculated for this curve.  The way you followed creates somewhat more work, but will finally achieve the same result.
This approach is somewhat similar to that of Clarinetist, but takes advantage of a particular feature of the parametric curve functions; it also proceeds entirely by implicit differentiation.  We may write $$ \ x \ = \ t^2 + 2t \  = \ (t+1)^2 - 1 \ \ , \ \ y \ = \ 3t^4 + 4t^3 \ = \ t^3 · (3t+4) \ = \ t^3 · [ (3t+3) + 1] \ \ . $$  We will then substitute $ \ u = t+1 \ \ ; $  consequently,
$$ x + 1 \ = \ u^2 \ \ \Rightarrow \ \ \frac{d}{dx} \ [x + 1] \ = \ \frac{d}{dx} \ [u^2] \ \ \Rightarrow \ \ 1 \ = \ 2u \ \frac{du}{dx} $$
$$ \Rightarrow \ \ \frac{d}{dx} \ [1] \ = \ \frac{d}{dx} \left[2u \ \frac{du}{dx} \right] \ \ \Rightarrow \ \ 0 \ = \ 2·\left(\frac{du}{dx} \right)^2 \ + \ 2u· \frac{d^2u}{dx^2} \ \ \Rightarrow \ \ \frac{d^2u}{dx^2} \ = \ -\frac{1}{u}·\left(\frac{du}{dx} \right)^2 \ \ . $$
For the other curve equation, we have
$$ y \ = \ t^3 · [ 3(t+1) + 1] \ \rightarrow \ (u-1)^3 · ( 3u + 1)  $$
$$ \Rightarrow \ \ \frac{dy}{du} \ = \ 3·(u-1)^2 · ( 3u + 1) \ + \ (u-1)^3 · 3 \ = \ 3·(u-1)^2 · [(3u+1) + (u-1)] $$ $$ = \ 12u·(u-1)^2 \ \ . $$
It follows that $$ \ \ \frac{dy}{dx} \ = \ \frac{dy}{du} · \frac{du}{dx} \ = \ 12u·(u-1)^2 \ · \ \frac{1}{2u} \ = \ 6·(u-1)^2 \ \rightarrow \ 6t^2 \ \ . $$
[We will remark upon this later.]
For the second derivative functions, we obtain
$$  \frac{d^2y}{du^2} \ = \  \frac{d}{du} \ [ 12u·(u-1)^2 ] \ = \ 12 ·(u-1)^2 \ + \ 12u · 2 \ (u-1) $$ $$ = \ 12 ·(u-1) · [(u-1) + 2u] \ = \ 12 ·(u-1) ·  (3u-1) \ \ ; $$
$$  \frac{d^2y}{dx^2} \ = \  \frac{d}{dx} \ \left[\frac{dy}{dx} \right] \ = \ \frac{d}{dx} \ \left[\frac{dy}{du} · \frac{du}{dx} \right] \ = \ \left(\frac{d}{du} \  \left[\frac{dy}{du} \right]  · \frac{du}{dx} \right) · \frac{du}{dx} \ + \ \frac{dy}{du} \ · \frac{d}{dx} \  \left[ \frac{du}{dx} \right] $$
$$  = \  \frac{d^2y}{du^2}  · \left(\frac{du}{dx} \right)^2  \ + \ \frac{dy}{du} \ ·  \frac{d^2u}{dx^2}   $$
[similar to your result, since $ \ \frac{du}{dx} = \frac{dt}{dx} ] $
(bringing in the expressions found above)
$$ = \  \frac{d^2y}{du^2}  · \left(\frac{du}{dx} \right)^2  \ + \ \frac{dy}{du} \ ·  \left[-\frac{1}{u}·\left(\frac{du}{dx} \right)^2 \right] \ = \ \left[\frac{d^2y}{du^2} \ - \ \frac{1}{u} · \frac{dy}{du} \right]  ·   \left(\frac{du}{dx} \right)^2    $$
$$  = \ \left[12·(u-1)·(3u-1) \ - \ \frac{1}{u} · 12u·(u-1)^2 \right]  ·   \left(\frac{1}{2u} \right)^2   $$
$$  = \ 12·(u-1) \ · \ [ (3u-1) \ - \  (u-1) ]  ·   \left(\frac{1}{2u} \right)^2 \ =  \ 12·(u-1) · 2u · \left(\frac{1}{2u} \right)^2 $$
$$ = \ \frac{6·(u-1)}{u} \ \ \rightarrow \ \ \frac{6t}{t \ +  \ 1} \ \ . $$
So this may not be the quickest route to the desired second derivative, but it is effective.
In the graphs at the start of this post, we see that this is a very narrow "hairpin" curve, with two "arms" almost entirely in the first quadrant, with a "turn" close to the origin [shown at lower right].  A curious feature about the curve (which is what interested me in this problem) is that the first derivative parametric function, $ \ 6t^2 \ , $ betrays no hint that there is something peculiar about the turn; the function is defined everywhere and is continuous, and the slope of the curve is always positive, except where it is zero at the origin.  However, the second derivative function is not defined at $ \ t = -1 \   $ (the point $ \ (-1,-1) \ ) , $ changes discontinuously from positive for $ \ t < -1 \ $ to negative for $ \ -1 < t < 0 \ , $ then reaches an inflection point at the origin $ \ (t=0) \ $ and returns to "upward" concavity for $ \ t > 0 \ \ . $
A: You do not have to use implicit differentiation. $x$ and $y$ are explicitly given in terms of parameter $t$.
$\displaystyle \frac{dy}{dt} = 12t^3 + 12t^2, \frac{dx}{dt} = 2t + 2$.
From the above, and chain rule, you get:  $\displaystyle \frac{dy}{dx} = \frac{dy}{dy} \cdot \frac{dt}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = 6t^2$, after simplification. Note that there is the assumption $t \neq -1$ made here to permit cancellation.
Now, $\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx}(\frac{dy}{dx}) = \frac{d}{dt}(\frac{dy}{dx}) \cdot \frac{dt}{dx} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}}$, again by chain rule.
Utilising results from before, $\displaystyle \frac{d^2y}{dx^2} = \frac{12t}{2t + 2} = \frac{6t}{t+1}$.
To find the correct value of $t$, solve for $x = 8 \implies t^2 + 2t = 8$, giving $t = 2$ or $t = -4$. By putting this into the expression for $y$, you note that only $t = 2$ gives the expected value of $y = 80$, so you need $t = 2$.
Putting that into the expression for the second derivative, you get $4$ as the answer.
A: Just for your information since you will need it quite often with differential equation problem
$$\large\frac{d^2 t}{dx^2}=-\frac {\frac{d^2 x}{dt^2} } {\Big[\frac{d x}{dt} \Big]^3}$$
Just checking
$$x=\tan(t) \implies \frac{d x}{dt}= \sec ^2(t)\implies \frac{d^2 x}{dt^2}=2 \tan (t) \sec ^2(t)\implies \frac{d^2 t}{dx^2}=-2 \sin (t) \cos ^3(t)$$
$$t=\tan^{-1}(x) \implies \frac{d t}{dx}=\frac{1}{1+x^2}\implies \frac{d^2 t}{dx^2}=-\frac{2 x}{\left(1+x^2\right)^2}=-2 \sin (t) \cos ^3(t)$$
