Editing Chebyshev polynomials (section)

===Differentiation and integration===
The derivatives of the polynomials can be less than straightforward. By differentiating the polynomials in their trigonometric forms, it can be shown that:
<math display="block">\begin{align}
\frac{\mathrm{d}T_n}{\mathrm{d}x} &= n U_{n - 1} \\
\frac{\mathrm{d}U_n}{\mathrm{d}x} &= \frac{(n + 1)T_{n + 1} - x U_n}{x^2 - 1} \\
\frac{\mathrm{d}^2 T_n}{\mathrm{d}x^2} &= n\, \frac{n T_n - x U_{n - 1}}{x^2 - 1} = n\, \frac{(n + 1)T_n - U_n}{x^2 - 1}.
\end{align}</math>

The last two formulas can be numerically troublesome due to the division by zero ({{Sfrac|0|0}} [[indeterminate form]], specifically) at {{math|1=''x'' = 1}} and {{math|1=''x'' = −1}}. By [[L'Hôpital's rule]]:
<math display="block">\begin{align}
\left. \frac{\mathrm{d}^2 T_n}{\mathrm{d}x^2} \right|_{x = 1} \!\! &= \frac{n^4 - n^2}{3}, \\
\left. \frac{\mathrm{d}^2 T_n}{\mathrm{d}x^2} \right|_{x = -1} \!\! &= (-1)^n \frac{n^4 - n^2}{3}.
\end{align}</math>

More generally,
<math display="block">\left.\frac{\mathrm{d}^p T_n}{\mathrm{d}x^p} \right|_{x = \pm 1} \!\! = (\pm 1)^{n+p}\prod_{k=0}^{p-1}\frac{n^2-k^2}{2k+1}~,</math>
which is of great use in the numerical solution of [[eigenvalue]] problems.

Also, we have:
<math display="block">\frac{\mathrm{d}^p}{\mathrm{d}x^p}\,T_n(x) = 2^p\,n\mathop{{\sum}'}_{0\leq k\leq n-p\atop k \,\equiv\, n-p \pmod 2}
\binom{\frac{n+p-k}{2}-1}{\frac{n-p-k}{2}}\frac{\left(\frac{n+p+k}{2}-1\right)!}{\left(\frac{n-p+k}{2}\right)!}\,T_k(x),~\qquad p \ge 1,</math>
where the prime at the summation symbols means that the term contributed by {{math|1=''k'' = 0}} is to be halved, if it appears.

Concerning integration, the first derivative of the {{mvar|T<sub>n</sub>}} implies that:
<math display="block">\int U_n\, \mathrm{d}x = \frac{T_{n + 1}}{n + 1}</math>
and the recurrence relation for the first kind polynomials involving derivatives establishes that for {{math|''n'' ≥ 2}}:
<math display="block">\int T_n\, \mathrm{d}x = \frac{1}{2}\,\left(\frac{T_{n + 1}}{n + 1} - \frac{T_{n - 1}}{n - 1}\right) = \frac{n\,T_{n + 1}}{n^2 - 1} - \frac{x\,T_n}{n - 1}.</math>

The last formula can be further manipulated to express the integral of {{mvar|T<sub>n</sub>}} as a function of Chebyshev polynomials of the first kind only:
<math display="block">\begin{align}
\int T_n\, \mathrm{d}x &= \frac{n}{n^2 - 1} T_{n + 1} - \frac{1}{n - 1} T_1 T_n \\
&= \frac{n}{n^2 - 1}\,T_{n + 1} - \frac{1}{2(n - 1)}\,(T_{n + 1} + T_{n - 1}) \\
&= \frac{1}{2(n + 1)}\,T_{n + 1} - \frac{1}{2(n - 1)}\,T_{n - 1}.
\end{align}</math>

Furthermore, we have:
<math display="block">\int_{-1}^1 T_n(x)\, \mathrm{d}x =
\begin{cases}
\frac{(-1)^n + 1}{1 - n^2} & \text{ if }~ n \ne 1 \\
0                          & \text{ if }~ n = 1.
\end{cases}</math>