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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?

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up vote 1 down vote accepted

Hint: $\frac{\sin x}x\to1$ when $x\to0$.

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