Second derivative of implicit equation in "How Not to Land at Lake Tahoe" I was trying to replicate the math "How not to land in Lake Tahoe" problem by Barshinger (1992) in the American Mathematical Monthly.
Essentially, he models the landing path of a plane with a cubic polynomial.

The resultant equation is
$y(x)=h({2(x/L)^3+3(x/L)^2})$
The velocity is intuitively the implicit differentiation of the landing path function as stated in the original problem solution:
$\frac{dy}{dt}=\frac{6Uh}{L}((x/L)^2+(x/L))$
$U = \frac{dx}{dt}$ assuming that the after the implicit differentiation, the horizontal velocity is constant.
However, I am stuck when he takes the second derivative to get the vertical acceleration of the plane on the landing path. I understand that in kinematics that acceleration is simply the derivative of the velocity. However, I am unsure how he arrived at the below expression for acceleration.
$\frac{d^2y}{dt^2}=\frac{6U^2h}{L^2}(2(x/L)+1)$
I understand the latter part - $(2(x/L)+1)$ which is just following the normal differentiation rules. However, I don't know how to arrive at the $\frac{6U^2h}{L^2}$ term with the $U^2$ and the $L^2$ components. Why are they raised to the power of 2? Is this some kind of chain rule or product rule that I'm missing?
 A: In this context, $U$, $h$ and $L$ are all constants and $x$ is a function of $t$: $x(t) = Ut$.
Now with $y'(t) = \frac{6Uh}{L}((x(t)/L)^2 + x(t)/L)$, you take the derivative again and get
$$\begin{align*}
y''(t) & = \frac{6Uh}{L}(2x(t)x'(t)/L^2 + x'(t)/L) \\
& = \frac{6Uh}{L}(2 x(t) U / L^2 + U/L) \\
& = \frac{6U^2h}{L^2}(2 x(t) + 1),
\end{align*}$$
so the only thing that is happening is that the constant factor $U/L$ appearing in both terms inside the parentheses is moved outside of the parentheses.
A: (A little too long for a comment)
It might be easier to see what is being done if common factors are extracted as soon as they appear. Starting with Barshinger's trajectory function, we have
$$ y \ \ = \ \ h·\left[ \ 2 \left(\frac{x}{L} \right)^3 \ + \ 3 \left( \frac{x}{L} \right)^2 \ \right] \ \ = \ \ \frac{h}{L^2}·\left[ \   \left(\frac{2}{L} \right)·x^3 \ + \ 3· x^2 \ \right] \ \ .  $$
So differentiation with respect to time produces
$$ \Rightarrow \ \ \frac{dy}{dt} \ \ = \ \ \frac{h}{L^2}·\left[ \   \left(\frac{6}{L} \right)·x^2 \ + \ 6· x \ \right]·\frac{dx}{dt} \ \ = \ \ \frac{6h}{L^2}·\left[ \   \left(\frac{1}{L} \right)·x^2 \ + \ x \ \right]·U $$
(with the Chain Rule producing the velocity factor, taken to be a constant value $ \ U \ \ ) $
$$ = \ \ \frac{6h·U}{L^2}·\left[ \   \left(\frac{1}{L} \right)·x^2 \ + \ x \ \right] \ \ . $$
The second implicit differentiation with respect to time in order to obtain the vertical component of acceleration creates another "chain rule factor" of horizontal velocity, giving us
$$ \Rightarrow \ \ \frac{d^2y}{dt^2} \ \ = \ \ \frac{6h·U}{L^2}·\left[ \   \left(\frac{2}{L} \right)·x \ + \ 1 \ \right]·\frac{dx}{dt} \ \ ,   $$
and so to the stated result.
