If $P_n(1)=1$ calculate $P'_n(1)$ in Legendre polynomials $P_n(x)$ is in $[-1,1]$ and $P_n(1)=1$ .The problem is getting $P'_n(1)$.
On Wikipedia it says that it is $\frac{n(n+1)}2$.
I derive the problem showed here
How could I prove that $P_n (1)=1=-1$ for the Legendre polynomials?
in order to get P'(n) but it didn't helped so much.
 A: The generating function for Legendre polynomials is
\begin{eqnarray*}
\sum_{n=0}^{\infty} P_n(z) h^n =\frac{1}{\sqrt{1-2hz+h^2}}. 
\end{eqnarray*}
Differentiate this w.r.t $z$
\begin{eqnarray*}
\sum_{n=0}^{\infty} P^{'}_n(z) h^n =\frac{(-1/2)(-2h)}{(1-2hz+h^2)^{3/2}}. 
\end{eqnarray*}
Set $z=1$
\begin{eqnarray*}
\sum_{n=0}^{\infty} \color{red}{P^{'}_n(1)} h^n =\frac{h}{(1-h)^{3}} =\sum_{n=0}^{\infty} \color{red}{\binom{n+1}{2}} h^n.
\end{eqnarray*}
A: Alternatively, it follows directly from Legendre's differential equation of 
$$(1 - x^2) y'' - 2xy' + n(n + 1) y = 0.\tag1$$
Since $y = P_n(x)$ is a solution to (1) we have
$$(1 - x^2) P''_n(x) - 2x P'_n (x) + n(n + 1) P_n (x) = 0.$$
Setting $x = 1$ in the above equation leads to
$$P'_n (1) = \frac{n(n + 1)}{2} P_n (1).$$
But since you already known that $P_n (1) = 1$, then the desired result immediately follows. 
A: It follows by induction from the recurrence
$$
 (n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x)
$$
Just differentiate both sides and use that $P_n(1)=1$. 
