<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] $x=t^2+2t$
$y=3t^4+4t^3$
$\displaystyle\frac{dx}{dt}=2t+2$ 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 ,
$\displaystyle\frac{d^2y}{dx^2}=\frac{6t}{t+1}=\frac{6*2}{2+1}=\fbox4$
</s>