<s>[INST] 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 $\frac{d^2 t}{dx^2}$?
 [/INST] 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.
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