Derivatives
The ladder relations have become:
$$
\left[ (1-x^2) \frac{d}{dx} - n\,x \right] T_n(x) = -n\; T_{n+1}(x) \\
\left[ (1-x^2) \frac{d}{dx} + n\,x \right] T_n(x) = +n\; T_{n-1}(x)
$$
If we add these equations togther again, then:
$$
2 (1-x^2) \frac{dT_n}{dx} = - n \left[ T_{n+1}(x) - T_{n-1}(x) \right]
$$
Giving for the derivative of a Chebyshev solution function:
$$
\frac{dT_n}{dx} = \frac{n}{2(x^2-1)} \left[T_{n+1}(x) - T_{n-1}(x)\right]
$$
Remember the recursion relationship:
$$
T_{n+1}(x) + T_{n-1}(x) = 2 x\,T_n(x)
$$
With help of this, we can do the following:
$$
T_{n+1}(x) - T_{n-1}(x) = 2 x\,T_n(x) - 2 T_{n-1}(x) =\\
2 x \left[ 2\,x T_{n-1}(x) - T_{n-2}(x) \right] - 2 T_{n-1}(x) =\\
4 (x^2-1) T_{n-1}(x) + 2 T_{n-1}(x) - 2 x\, T_{n-2}(x) =\\
4 (x^2-1) T_{n-1}(x) + 2 T_{n-1}(x) - \left[ T_{n-1}(x) + T_{n-3}(x) \right] =\\
4 (x^2-1) T_{n-1}(x) + \left[ T_{n-1}(x) - T_{n-3}(x) \right]
$$
Herewith:
$$
\frac{dT_n}{dx} = \frac{n}{2(x^2-1)} \left[T_{n+1}(x) - T_{n-1}(x)\right] =\\
2 n T_{n-1}(x) + \frac{n}{2(x^2-1)} \left[T_{n-1}(x) - T_{n-3}(x)\right]
$$
So we have a recursion formula expressing the derivative of Chebyshev function
into a sequence of lower order Chebyshev functions:
$$
\frac{dT_n}{dx} = 2 n T_{n-1}(x) + 2 n T_{n-3}(x) + 2 n T_{n-5}(x) +
\; .\,.\,. \; + \; ( \mbox{last term} )
$$
Where the last term can be one of these two:
$$
\frac{n}{2(x^2-1)} \left[T_3(x) - T_1(x)\right] \quad \vee \quad
\frac{n}{2(x^2-1)} \left[T_2(x) - T_0(x)\right]
$$
If solutions of the Chebyshev differential equation are restricted to
Chebyshev polynomials of the first kind, then we have
$T_3(x) = 4x^3-3x\;$, $T_2(x) = 2x^2-1\;$,
$T_1(x) = x\;$, $T_0(x) = 1\;$. Herewith:
$$
\frac{n}{2(x^2-1)} \left[T_3(x) - T_1(x)\right] = 2 n . x \quad \vee \quad
\frac{n}{2(x^2-1)} \left[T_2(x) - T_0(x)\right] = n
$$
Let's concentrate on the value of the derivative at $x=1$. From the definition
of the Chebyshev polynomials $T_n(x) = \mbox{cos[h]}(n\,\mbox{arccos[h]}(x))$
it is clear that, for all $n$ : $T_n(1) = 1$. And $x=1$ is the only place where
all polynomials assume that maximum value, at the interval $\left[ -1,+1
\right]$. Thus the derivatives $dT_n(x)/dx$ assume their maximum value at that
point as well. We can even calculate what the maximum is.
For even $n$ we have $n/2$ terms $2n$, giving a total of $n/2.2n = n^2$.
For odd $n$ we have $(n-1)/2$ terms equal to $2n$ and one term equal to $n$,
giving a total of $(n-1)/2.2n+n = n^2$. Both sequences add up to $n^2$. Thus:
$$
\left. \frac{dT_n(x)}{dx} \right|_{x=1} = n^2
$$