Third derivative of $y=at^2+2bt+c$ and $t=ax^2+bx+c$ If $y=at^2+2bt+c$ and $t=ax^2+bx+c$. Then find
$$\frac{d^3y}{dx^3}$$
Now $\frac{dy}{dx}=(2at+2b).(2ax+2b)$ but to proceed further as  $\frac{dy}{dx}$ is function of $x,t$
 A: Hint: Replace $t$ by $ax^2+bx+c$ in $at^2+2bt+c$, you will obtain $y(x)$  and calculate the third derivative of the function $y(x)$.
A: Plugging in $t$ is an unsatisfying way to do it. You can just repeatedly apply the chain rule. I'll find the second derivative for you. To get to the third, it is similar work.
\begin{align*}
\frac{dy}{dx} &= \frac{dy}{dt} \frac{dt}{dx}\\
\frac{d^2y}{d^2x} &= \frac{d}{dx}\left(\frac{dy}{dx}\right)\\
&= \frac{d}{dx}\left(\frac{dy}{dt} \frac{dt}{dx}\right)\\
&=\left(\frac{d}{dx}\left(\frac{dy}{dt}\right)\right)\frac{dt}{dx} + \frac{dy}{dt}\left(\frac{d}{dx}\left(\frac{dt}{dx}\right)\right)\\
&= \left(\frac{d^2y}{d^2t}\frac{dt}{dx}\right)\frac{dt}{dx} + \frac{dy}{dt}\left(\frac{d^2t}{dx^2}\right)\\
&= \frac{d^2y}{d^2t}\left(\frac{dt}{dx}\right)^2 + \frac{dy}{dt}\left(\frac{d^2t}{dx^2}\right)
\end{align*}
A: There are two approaches. First, as Tsemo wrote, you can simply plug in $t(x)$ into $y$ and then differentiate. The second approach, suggested by Angel, is to repeatedly apply the chain rule:
$$\frac{d}{dx} = \frac{\partial}{\partial x} + \frac{dt}{dx}\frac{\partial}{\partial t}$$
so that
$$\frac{dy}{dx} = (2at+b)(2ax+b)$$
$$\frac{d^2y}{dx^2} = (2at+b)(2a) + 2a(2ax+b)^2$$
$$\frac{d^3y}{dx^3} = 8a^2(2ax+b) + 4a^2(2ax+b) = 24a^3x+12a^2b.$$
The advantage of this approach is that it will continue to work even if the relation coupling $t$ and $x$ is complicated so that $t$ cannot easily be written as a function of $x$.
