Rodrigues formula for Chebyshev rational functions. Could you calculate a Rodrigues formula for the Chebyshev rational functions defined by
$$T_n\left(\frac{x-1}{x+1}\right)$$
where $T_n$ is a Chebyshev polynomial.
 A: This is probably not the sort of formula you are looking for but let me write it for reference of how Rodrigues formulas for the classic polynomials are computed. Perhaps, perhaps not, the procedure can be adapted for these rational functions.
We have a Rodrigues formula for Chebyshev polynomials. Let us deduce it.
Chebyshev polynomials are (a particular case of the Jacobi polynomials) orthogonal with respect to
$$\int_{-1}^{1}T_m(x)T_n(x)K(x)dx$$
where $K(x)=(x^2-1)^{1/2}$.
Let us do the following (integrate by parts many times)
$$\begin{align}\int_{-1}^{1}\left((x^2-1)^nK(x)\right)^{(n)}x^mdx&=-\int_{-1}^{1}\left((x^2-1)^nK(x)\right)^{(n-1)}mx^{m-1}dx\\&=...=(-1)^mm!\int_{-1}^{1}\left((x^2-1)^nK(x)\right)^{(n-m)}dx\\&=(-1)^mm!\left((x^2-1)^nK(x)\right)^{(n-m-1)}|_{-1}^{1}\\&=0\end{align}$$
This means that 
$$\int_{-1}^{1}\frac{1}{K(x)}\left((x^2-1)K(x)\right)^{(n)}x^mK(x)dx=0$$
This means that $\frac{1}{K(x)}\left((x^2-1)K(x)\right)^{(n)}$ is orthogonal to all powers $m<n$. This implies that it is an orthogonal system with respect to the product above. By uniqueness, if we adjust the leading coefficient we should get the Chebyshev polynomials. From this it follows the formula

$$T_n(x)=(x^2-1)^{-1/2}\frac{1}{2^nn!}\frac{d^n}{dx^n}\left((x^2-1)^n(x^2-1)^{1/2}\right)$$

You could compose with $\frac{x-1}{x+1}$ to get a formula for $T_n\left(\frac{x-1}{x+1}\right)$ but this composition would appear after the derivatives. This is not the way Rodrigues formulas look like.
A: Let we just follow the approach in the paragraph Rodrigues' formula here. By setting
$$ A_n(x)=T_n\left(\frac{x-1}{x+1}\right)\tag{1} $$
and $x=\frac{1+\cos\theta}{1-\cos\theta}=\cot^2\frac{\theta}{2}$ we have:
$$ A_n\left(\frac{1+\cos\theta}{1-\cos\theta}\right) = \cos(n\theta)\tag{2} $$
and by differentiating two times both sides with respect to $\theta$ we get:
$$2(2+\cos\theta)\,A_n'(x)+4x\, A_n''(x) = -n^2 \cos(n\theta) = -n^2 A_n(x)$$
or:
$$ 4x\,A_n''+2\frac{1+3x}{1+x}\,A_n'+n^2 A_n = 0\tag{3} $$
and the associated functions with this Sturm-Liouville-type differential equation are $R(x)=(1+x)\sqrt{x}$ and $W(x)=\frac{1+x}{4\sqrt{x}}$. This gives that $A_n$ should be proportional to:
$$\frac{1}{W(x)}\frac{d^n}{dx^n}\left(W(x)\cdot(4x)^n\right)$$
or to:
$$ \frac{4^n\sqrt{x}}{1+x}\cdot\frac{d^n}{dx^n}\left((1+x)\cdot x^{n-\frac{1}{2}}\right).\tag{4}$$
However, we can notice that the condition on the coefficient of $A_n'$ in $(3)$ to be linear is not met, so, if $(4)$ is not the Rodrigues representation for $A_n$, this set of functions does not have a Rodrigues representation.
