<s>[INST] GRE Practice Question I need a calculus refresher! This question was one of the lowest percent correct on the practice (27%). 
A curve in the xy-plane is given by
\begin{align*}
x &= t^2+2t \\
y &= 3t^4+4t^3
\end{align*}
for all $t > 0.$ The value of $\frac{d^2y}{dx^2}$ at the point $(8,80)$ is
(a) 4, (b) 24, (c) 32, (d) 96, (e) 192
The first thing we can do is figure out what the $t-$value is at the point $(8,80)$. Using the first equation we have $8 = t^2 + 2t.$ By solving the quadratic we get $t = 2$ and $t = -4$, but only the first solution is in the domain. We can check to see that $t = 2$ satisfies the second equation.
Then comes the part I'm not sure about. Can someone explain how you determine the derivative (first and second) of $x$ with respect to $y$? I think I remember $\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt}$ because it sort of cancels like a fraction. 
 [/INST] Solve the quadratic equation $t^2+2t-x=0$
 to find $t=-1+\sqrt{1+x}$ and substitute into $y$ to get:
$$y=3(-1+\sqrt{1+x})^4+4(-1+\sqrt{1+x})^3.$$
Thus:
$$y'=12(-1+\sqrt{1+x})^3\cdot \frac{1}{2\sqrt{1+x}}+12(-1+\sqrt{1+x})^2\cdot \frac{1}{2\sqrt{1+x}}.$$
$$y''=-\frac{6}{2(1+x)^{3/2}}\cdot((-1+\sqrt{1+x})^3+(-1+\sqrt{1+x})^2)+\frac{6}{\sqrt{1+x}}\cdot (3(-1+\sqrt{1+x})^2\cdot \frac{1}{2\sqrt{1+x}}+2(-1+\sqrt{1+x})\cdot \frac{1}{2\sqrt{1+x}}).$$
$$y''(8)=-\frac{6}{2\cdot 27}(8+4)+2(2+\frac{2}{3})=4.$$
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