I'm trying to show that the derivative of Chebyshev polynomials at $x = 1$ satisfy $T_k'(1) = k^2$ for each $k ≥ 0$.
I can get the derivative to come out as
$T'_k(x) = \frac{k \sin(k\theta)}{\sin(\theta)} $
but after that it always ends up as just zero and not $k^2$....
What am I missing?