This is a question related to the range-direction-cosine coordinate system, in case that helps. There is a unique property for this system in which one of the coordinate values, $w$, can be deduced from two others, $u$ and $v$.
Let a function $w(t) = \sqrt{1 - u(t)^2 - v(t)^2}$. I am hoping to find a way of expressing the $\text{n}^{\text{th}}$ derivative programmatically in terms of the various derivatives of $u(t)$ and $v(t)$.
I have put found the first 5 derivatives using Wolfram but unfortunately have been as yet unable to find the pattern. This was simple to do for something like $\frac{d}{dt}\left(f(t) \cdot g(t)\right)$ where each derivative is just based on Pascal's triangle.
Any suggestions on a good approach, if not necessarily the answer, would be appreciated.
You can use the technique of implicit differentiation to make this problem a little easier. For the moment, I will suppress the $t$ term and simply write $w = w(t)$, $u = u(t)$ and $v = v(t)$. We start with
$$w = \sqrt{1 - u^2 - v^2}$$
After some short algebra, we obtain the equivalent form
$$w^2 + u^2 + v^2 = 1.$$
Now imagine deriving through both sides with respect to $t$. On the right, $\frac{d}{dt}(1) = 0$ of course. On the left, each of the squared terms can be derived using the chain rule. For example,
$$\frac{d}{dt}(w^2) = 2w \cdot\frac{d}{dt}(w) = 2w\frac{dw}{dt}$$
Likewise for the $u$ and $v$ terms. So we obtain $2ww' + 2uu' + 2vv' = 0$, or (after dividing the $2$ away) just $ww' + uu' + vv' = 0$.
From this equation, it is possible to isolate $w'$ and then substitute in $w = \sqrt{1 - u^2 - v^2}$ to get the final answer for $w'$ in terms of $u$ and $v$ (and $u'$ and $v'$).
For higher order derivatives, you can start from $ww' + uu' + vv' = 0$ and use your formula for $\frac{d^n}{dt^n}(f(t)g(t))$ to derive this equation $n$ more times, with $f = w$ and $g = w'$ for the first term, and so on for the $u$ and $v$ terms.
However, the result is a polynomial in $w$, $w'$, $w''$, $\ldots$ which may be impossible to solve in closed form. You could probably get a recursive formula for $w^{(n)}$ (written in terms of $u$, $v$, and $w^{(n-1)}$), but you might not be able to do any better than that.
Edit: Since you mentioned a recursive formula would be acceptable, here is a derivation of precisely that. We begin with $ww' = - uu' - vv'$ and, assuming $w, w', w^{(2)}, \ldots, w^{(n)}$ are known, produce a formula for $w^{(n+1)}$.
We have the following generalized product rule.
$$\frac{d^n}{dt^n}\big(f(t)g(t)\big) = \sum_{k=0}^n \binom{n}{k} f^{(k)}(t) g^{(n-k)}(t)$$
Choosing $f(t) = w(t)$ and $g(t) = w'(t)$, we see
$$\frac{d^n}{dt^n}(ww') = \sum_{k=0}^n \binom{n}{k} w^{(k)} w^{(n-k+1)}$$
Expanded to a few terms, this sum is
$$\binom{n}{0}w^{(0)}w^{(n+1)} + \binom{n}{1}w^{(1)}w^{(n)} + \binom{n}{2}w^{(2)}w^{(n-1)} + \cdots$$
Recall $w^{(0)} = w$ and $\binom{n}{0} = 1$. Notice in particular that, because $k$ ranges from $0$ to $n$, only one term in the above sum (namely, the first term) contains $w^{(n+1)}$ as a factor.
So, if we begin with $ww' = - uu' - vv'$ and distribute $\frac{d^n}{dt^n}$ throughout, we may isolate the term $w^{(n+1)}$ and solve for it as follows:
$$w^{(n+1)} = -\frac{1}{w}\Big(uu^{(n+1)} + vv^{(n+1)} + \sum_{k=1}^n \binom{n}{k} \big(w^{(k)}w^{(n-k+1)} +u^{(k)}u^{(n-k+1)} + v^{(k)}v^{(n-k+1)}\big)\Big)$$
Note I removed the $k = 0$ term from the sums for $uu'$ and $vv'$, so that the remaining sum component now begins from $k = 1$. Note also, every derivative of $w$ appearing in the LHS is at least $0$ and at most $n$, so this is indeed a proper recursive formula for $w^{(n+1)}$.