A general formula for the $n$-th derivative of a parametrically defined function Letting
$$\begin{align*}x&=f(t)\\y&=g(t)\end{align*}$$
the following expressions are well known:
$$\begin{align*}\frac{\mathrm dy}{\mathrm dx}&=\frac{g^\prime (t)}{f^\prime (t)}\\\frac{\mathrm d^2 y}{\mathrm dx^2}&=\frac{f^\prime (t)g^{\prime\prime} (t)-g^\prime (t)f^{\prime\prime} (t)}{f^\prime (t)^3}\end{align*}$$
With some effort, we can derive the expression for the third derivative:
$$\frac{\mathrm d^3 y}{\mathrm dx^3}=\frac{f'(t) \left(g^{(3)}(t) f'(t)-3 f''(t) g''(t)\right)+g'(t) \left(3 f''(t)^2-f^{(3)}(t)f'(t)\right)}{f'(t)^5}$$
After deriving expressions for the next higher derivatives, I am unable to detect any particular pattern in the expressions, save for the denominator $f'(t)^{2n-1}$ of the $n$-th derivative. I've also tried to search around for information on the derivatives of parametrically-defined functions, but no dice.
Here then is my question: is there a general formula for $\dfrac{\mathrm d^n y}{\mathrm dx^n}$ in terms of $f(t),g(t)$ and their derivatives?
 A: I am not sure if what follows is useful for symbolic manipulations, but it is definitely useful for numerical evaluation of these derivatives.
First, let me state the "closed form" expression for $n^\text{th}$ derivative of $f^{-1}(x)$. Naturally, it is found by fiddling with $f(f^{-1}(x))=x$. Since the above is a composition, it is natural to expect appearance of Bell polynomials (see ref. by Paolo Ricci).
The derivation has been done by a trial and error:
$$
    \frac{\mathrm{d}^n}{\mathrm{d} x^n} f^{-1}(x) = Y\left(\begin{array}{cc} 
    \frac{(n)_1}{ f^{-1}(x)} & -\frac{ f^{(2)}( f^{-1}(x)) }{ 2 \left(f^{-1}(x)\right)^2  } \\
    \frac{(n)_2}{ f^{-1}(x)} & -\frac{ f^{(3)}( f^{-1}(x)) }{ 3 \left(f^{-1}(x)\right)^3  } \\
     \vdots & \vdots \\
    \frac{(n)_{n-1}}{ f^{-1}(x)} & -\frac{ f^{(n)}( f^{-1}(x)) }{ n \left(f^{-1}(x)\right)^n  }
\end{array} \right)
  \qquad \qquad \text{for} \quad n \ge 2
$$
Or in Mathematica:
In[136]:= Table[
 With[{z = InverseFunction[f][x]}, 
  BellY[Table[{Pochhammer[n, j]/
       Derivative[1][f][
        z], -Derivative[j + 1][f][
         z]/((j + 1) Derivative[1][f][z]^(j + 1))}, {j, n - 1}]] - 
    D[z, {x, n}] // Simplify], {n, 2, 6}]

Out[136]= {0, 0, 0, 0, 0}

Now, to answer your question, we write $y = g(f^{-1}(x))$, and apply another Bell polynomial for derivative of a composition:
In[137]:= 
With[{p = 4, z = InverseFunction[f][x]}, 
  Sum[Derivative[m][g][z] BellY[p, m, 
      Table[If[k == 1, 1/Derivative[1][f][z], 
        BellY[Table[{Pochhammer[k, j]/
           Derivative[1][f][
            z], -Derivative[j + 1][f][
             z]/((j + 1) Derivative[1][f][z]^(j + 1))}, {j, 
           k - 1}]]], {k, p}]], {m, 1, p}] - (D[
     g[z], {x, p}])] // Expand

Out[137]= 0

