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.

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. 
