Parametric Derivatives

What is the nth derivative of a parametrically-defined function?  For the nth derivative of the product of two functions there is Leibniz’s rule, and for the nth derivative of the composition of two functions there is Faà di Bruno’s formula.  For x = f(t), y = g(t), though, what is \displaystyle \frac{d^n y}{d x^n}?  As far as I know there is no general formula.  In this post I will talk about some of the difficulties and consider a special case that I think gives insight into the general problem.  (This post was inspired by a recent math.SE question by user J.M.)

Via the chain rule we have \displaystyle \frac{dy}{dt} = \frac{dy}{dx} \frac{dx}{dt}, so \displaystyle \frac{dy}{dx} = \frac{g'(t)}{f'(t)}.

Similarly, if we denote \displaystyle \frac{dy}{dx} by y', we have \displaystyle \frac{d y'}{dt} = \frac{d y'}{dx} \frac{dx}{dt}, so \displaystyle \frac{d^2 y}{d x^2} = f'(t)^{-1} \frac{d}{dt} \left(\frac{dy}{dx}\right) = \frac{f'(t)g''(t) - f''(t)g'(t)}{(f'(t))^3}.

In general, we can see that \displaystyle \frac{d^n y}{d x^n} takes the form \displaystyle \left(\frac{1}{f'(t)} \frac{d}{dt}\right)^n g(t).  Thus the difficulty in finding a nice expression lies with f(t); the uglier the form of f'(t), the nastier \displaystyle \frac{d^n y}{d x^n} will be.

So let’s look at a special case where f'(t) is complicated enough that we get something interesting but not so complicated that the result is intractable.  Taking f'(t) = 1 just gives us \displaystyle \frac{d^n y}{d x^n} = g^{(n)}(t), so let’s try f'(t) = t.  (Thus we have, say, f(t) = t^2/2.)

Calculating the first few derivatives by hand yields
\displaystyle \frac{dy}{dx} = \frac{g'(t)}{t}, \\    \frac{d^2y}{dx^2} = \frac{g''(t)t - g'(t)}{t^3}, \\    \frac{d^3y}{dx^3} = \frac{g'''(t)t^2 - 3g''(t)t + 3g'(t)}{t^5}, \\    \frac{d^4y}{dx^4} = \frac{g^{(4)}(t)t^3 - 6 g'''(t)t^2 + 15g''(t)t - 15g'(t)}{t^7}.

It looks like we’re getting a triangle of coefficients in the numerator, while the denominator is of the form t^{2n-1}.  Looking up the coefficients in the OEIS, it appears these are the coefficients of the reverse Bessel polynomials \phi_n.  In fact, this is the case.  Since t^k is always paired with g^{(k+1)}(t), and the signs alternate with the leading coefficient always being 1, the precise statement is
\displaystyle \frac{d^n y}{dx^n} = \frac{(-1)^{n-1} \phi_{n-1} (-t D(g'))}{t^{2n-1}},
where D is the differentiation operator.  (Thus D^k(g') = g^{(k+1)}(t), and D^0 (g') = g'(t).)  I prove this in my answer to J.M.’s question.  The derivation is kind of long, and so I’m not going to reproduce it here.

What can we conclude about the general case from this?  We’ll have (f'(t))^{2n-1} in the denominator.  In the numerator, the argument goes through to show that the coefficients of the terms containing only powers of f' and f'' and derivatives of g are those from the reverse Bessel polynomials.  Characterizing the other terms remains to be done, and that appears (to me) to be difficult.

In some respects it’s not surprising that the Bessel polynomials show up here.  After all, the Bessel functions j_n, y_n, i^{(1)}_n, i^{(2)}_n, k_n satisfy Rayleigh’s formulas:
\displaystyle j_n(t) = t^n \left( - \frac{1}{t} \frac{d}{dt} \right)^n \frac{\sin t}{t}, \\y_n(t) = -t^n \left( -\frac{1}{t} \frac{d}{dt} \right)^n \frac{\cos t}{t}, \\    i^{(1)}_n(t) = t^n \left( \frac{1}{t} \frac{d}{dt} \right)^n \frac{\sinh t}{t}, \\    i^{(2)}_n(t) = t^n \left( \frac{1}{t} \frac{d}{dt} \right)^n \frac{\cosh t}{t}, \\    k_n(t) = (-1)^n t^n \left( \frac{1}{t} \frac{d}{dt} \right)^n \frac{\pi e^{-t}}{2 t},
which are special cases of the formula above for \displaystyle \frac{d^n y}{dx^n}.

Added (12-20-2012): Todorov’s paper “New Explicit Formulas for the nth Derivative of Composite Functions” (Pacific Journal of Mathematics 92 (1): 1981, pp. 217-236) addresses this question as well.  I’ve added a summary of Todorov’s main results with respect to parametric derivatives to my answer to J.M.’s question on math.SE.

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