I need a little help to find the recurrence relation $$\sqrt{1-x^2}P_l^m(x) = \frac{1}{2l+1} (P_{l-1}^{m+1}-p_{l+1}^{m+1})$$ Using the identity $$(2l+1)P_l(x) = \frac{d}{dx}(P_{l+1}(x)-P_{l-1}(x))$$
I have $$P_l^m(x) = (-1)^m(1-x^2)^{m/2}\frac{d^m}{dx}P_l(x)$$
I guess I have to isolate $P_l(x)$, but I'm not sure how to manipulate this expression to plug it in the identity. I'm really stuck, any help will be appreciate.
For the Legendre polynomials, the difference $$ P_{l + 1}(x) - P_{l - 1}(x) $$ satisfies the identity $$ (2 l + 1) P_l(x) = {d \over dx}\left( P_{l + 1}(x) - P_{l - 1}(x) \right) \tag{1} $$
If we act on the identity (1) with $$ (-1)^m (1 - x^2)^{(m + 1)/2} d^m, $$ the result follows.