= Solution
The $n$ roots of $T_n$ are distinct and lie in $(-1,1)$. Repeated use of Rolle's theorem shows that every root of $T_n^{(k)}$, for $0\leq k<n$, also lies in $(-1,1)$. Its leading coefficient is positive, so
$$
T_n^{(k)}(x)>0\qquad(x\geq1).
$$
Taking $k=r+1$ shows that
$$
\frac d{dx}T_n^{(r)}(x)=T_n^{(r+1)}(x)>0
$$
for $0\leq r\leq n-1$; when $r=n-1$, the <derivative> is the positive constant $T_n^{(n)}$. This is the <positivity of Chebyshev derivatives beyond the unit interval>.
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