<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] 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
$$x=\tan(t) \implies \frac{d x}{dt}= \sec ^2(t)\implies \frac{d^2 x}{dt^2}=2 \tan (t) \sec ^2(t)\implies \frac{d^2 t}{dx^2}=-2 \sin (t) \cos ^3(t)$$
$$t=\tan^{-1}(x) \implies \frac{d t}{dx}=\frac{1}{1+x^2}\implies \frac{d^2 t}{dx^2}=-\frac{2 x}{\left(1+x^2\right)^2}=-2 \sin (t) \cos ^3(t)$$
</s>