Generalized Product Rule D^(n-m) (x^2 -1)^n (LaTeX inside)

[PhDeezNutz]

Homework Statement

Show: $D^{(n-m)} (x^2-1)^n = \frac{(n-m)!}{(n+m)!} (x^2-1)^m D^{(n+m)} (x^2-1)^n$

Hint: $D^{(n-m)} (x^2-1)^n = D^{(n-m)} [(x-1)^n (x+1)^n]$

Homework Equations

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Leibniz Rule for Differentiation:

$$D^k (uv) = \sum_{j=0}^k \binom{k}{j} D^j (u) D^{(k-j)} (v)$$

The Attempt at a Solution

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$$D^{(n-m)}[(x-1)^n (x+1)^n] = \sum_{j=0}^{n-m} \binom{n-m}{j} D^j (x-1)^n D^{(n-m-j)} (x+1)^n$$

$$D^{(n-m)}[(x-1)^n (x+1)^n] = \sum_{j=0}^{n-m} \frac{(n-m)!}{j! (n-m-j)!} D^j (x-1)^n D^{(n-m-j)} (x+1)^n$$

I don't know where to go from here. Somehow we're supposed to get rid of the sum by imposing conditions on the summing index such as $j \leq n$ but I'm still not getting it.

Any help would be appreciated. Thanks in advance.

 

[fresh_42]

What will you get, if you differentiate $x^n\,j-$times? Does it differ from $D^j(x+c)^n\,$?