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. 
 A: Note $$\frac{dy}{dx} = \frac{dy(t)}{dt} \left(\frac{dx(t)}{dt}\right)^{-1} = \frac{12t^3 + 12t^2}{2t+2} = 6t^2.$$  Note we can regard $t$ as an implicit function of $x$.  Then by the chain rule $$\frac{d^2y}{dx^2} = \frac{d}{dx}\left[6t(x)^2\right] = 12t(x)t'(x).$$  Since $(x,y) = (8,80)$ corresponds to $t = 2$, and $$t'(x) = \frac{1}{x'(t)} = \frac{1}{2(t+1)} = \frac{1}{6},$$  we obtain $$\frac{d^2 y}{dx^2} = 12(2)(1/6) = 4.$$
A: $$\begin{align*}
x' &= 2t+2 \\
y' &= 12t^3+12t^2
\end{align*}$$
The first derivative is 
$$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{12t^3+12t^2}{2t+2}=6t^2$$
For the second derivative you must use
$$\frac{d^2y}{dx^2}=\frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{{\frac{dx}{dt}}}=\frac{12t}{2t+2}=\frac{6t}{t+1}$$
for $t=2$ we get $\dfrac{d^2y}{dx^2}=\dfrac{12}{3}=4$
Hope this helps
A: 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:
$$y'=12(-1+\sqrt{1+x})^3\cdot \frac{1}{2\sqrt{1+x}}+12(-1+\sqrt{1+x})^2\cdot \frac{1}{2\sqrt{1+x}}.$$
$$y''=-\frac{6}{2(1+x)^{3/2}}\cdot((-1+\sqrt{1+x})^3+(-1+\sqrt{1+x})^2)+\frac{6}{\sqrt{1+x}}\cdot (3(-1+\sqrt{1+x})^2\cdot \frac{1}{2\sqrt{1+x}}+2(-1+\sqrt{1+x})\cdot \frac{1}{2\sqrt{1+x}}).$$
$$y''(8)=-\frac{6}{2\cdot 27}(8+4)+2(2+\frac{2}{3})=4.$$
