How to see there exists const $C$ such that $\frac{d^{n-m}}{dx^{n-m}}(1-x^2)^n=C(1-x^2)^m\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^n$ This comes up in relation to Legendre functions. The claim is made that for $n =0,1,2,3,\cdots$ and $m=0,1,2,3,\cdots,n$, there is a constant $C_{n,m}$ such that
$$
     \frac{d^{n-m}}{dx^{n-m}}(1-x^2)^n=C_{n,m}(1-x^2)^{m}\frac{d^{n+m}}{dx^{n+m}}(1-x^2)^n
$$
A typical suggested proof is writing out all of the polynomial coefficients. Does anyone have a clever way to see this must be true? I've tried many things with differentiation, but I can't find any clean way to see this.
Example: Consider $n=3,m=1$ where $(1-x^2)^n=1-3x^2+3x^4-x^6$, and
\begin{align}
    \frac{d^{3-1}}{dx^{3-1}}(1-x^2)^3&=-6+36x^2-30x^4 \\
    (1-x^2)^1\frac{d^4}{dx^4}(1-x^2)^3&=(1-x^2)(72-360x^2) \\
     &=72-72x^2-360x^2+360x^4 \\
     &=72-432x^2+360x^4 \\
     &=(-12)(-6+36x^2-30x^4)
\end{align}

Suppose we seek to determine the constant $Q$ in the equality
$$ Q_{n,m} \left(\frac{d}{dz}\right)^{n-m} (1-z^2)^n
= (1-z^2)^m \left(\frac{d}{dz}\right)^{n+m}
(1-z^2)^n$$
where $n\ge m.$ We will compute the coefficients on $[z^q]$ on the LHS
and the RHS. Writing $1-z^2 = (1+z)(1-z)$ we get for the LHS
$$\sum_{p=0}^{n-m} {n-m\choose p} 
{n\choose p} p! (1+z)^{n-p} 
\\ \times {n\choose n-m-p} (n-m-p)! (-1)^{n-m-p} (1-z)^{m+p}
\\ = (n-m)! (-1)^{n-m}
\sum_{p=0}^{n-m} {n\choose p} {n\choose n-m-p}
(1+z)^{n-p} (-1)^p (1-z)^{m+p}.$$
Extracting the coefficient we get
$$(n-m)! (-1)^{n-m}
\sum_{p=0}^{n-m} {n\choose p} {n\choose n-m-p} (-1)^p 
\\ \times \sum_{k=0}^{n-p} {n-p\choose k} 
(-1)^{q-k} {m+p\choose q-k}.$$
We use the same procedure on the RHS and merge in the $(1-z^2)^m$ term
to get
$$(n+m)! (-1)^{n+m}
\sum_{p=0}^{n+m} {n\choose p} {n\choose n+m-p} (-1)^p 
\\ \times \sum_{k=0}^{n+m-p} {n+m-p\choose k} 
(-1)^{q-k} {p\choose q-k}.$$
Working in parallel with LHS and RHS we treat the inner sum of the LHS
first, putting
$${m+p\choose q-k} = 
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q-k+1}} (1+z)^{m+p}
\; dz$$
to get
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m+p}
\sum_{k=0}^{n-p} {n-p\choose k} (-1)^{q-k} z^k \; dz
\\ = \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m+p} (1-z)^{n-p}
\; dz.$$
Adapt and repeat to obtain for the inner sum of the RHS
$$\frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{p} (1-z)^{n+m-p}
\; dz.$$
Moving on to the two outer sums we introduce
$${n\choose n-m-p} = 
\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m-p+1}} (1+w)^n
\; dw$$
to obtain for the LHS
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m} (1-z)^{n}
\sum_{p=0}^{n-m} {n\choose p} (-1)^p w^p \frac{(1+z)^p}{(1-z)^p}
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m} (1-z)^{n}
\left(1-w\frac{1+z}{1-z}\right)^n
\; dz\; dw
\\ = \frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n-m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1+z)^{m}
(1-z-w-wz)^n
\; dz\; dw.$$
Repeat for the RHS to get
$$\frac{1}{2\pi i}
\int_{|w|=\gamma}
\frac{1}{w^{n+m+1}} (1+w)^n
\\ \times \frac{(-1)^q}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{q+1}} (1-z)^{m}
(1-z-w-wz)^n
\; dz\; dw.$$
Extracting coefficients from the first integral (LHS) we write
$$(1-z-w-wz)^n = (2-(1+z)(1+w))^n
\\ = \sum_{k=0}^n {n\choose k} (-1)^k (1+z)^k (1+w)^k 2^{n-k}$$
and the inner integral yields
$$(-1)^q \sum_{k=0}^n {n\choose k} (-1)^k {m+k\choose q} 
(1+w)^k 2^{n-k}$$
followed by the outer one which gives
$$(-1)^q \sum_{k=0}^n {n\choose k} (-1)^k {m+k\choose q} 
{n+k\choose n-m} 2^{n-k}.$$
For the second integral (RHS) we write
$$(1-z-w-wz)^n = ((1-z)(1+w)-2w)^n
\\ = \sum_{k=0}^n {n\choose k} (1-z)^k (1+w)^k 
(-1)^{n-k} 2^{n-k} w^{n-k}$$
and the inner integral yields
$$(-1)^q \sum_{k=0}^n {n\choose k} {m+k\choose q} (-1)^q
 (1+w)^k (-1)^{n-k} 2^{n-k} w^{n-k}$$
followed by the outer one which produces
$$\sum_{k=0}^n {n\choose k} {m+k\choose q}
{n+k\choose k+m} (-1)^{n-k} 2^{n-k}.$$
The two sums are equal up to a sign and the RHS for the coefficient on
$[z^q]$ is obtained from the LHS by multiplying by 
$$\frac{(n+m)!}{(n-m)!} (-1)^{n-q}.$$
Observe that  powers of $z$  that are present  in the LHS and  the RHS
always  have the same  parity, the  coefficients being  zero otherwise
(either all even powers or all odd). Therefore $(-1)^{n-q}$ is in fact
a constant not  dependent on $q$, the question  is which.  The leading
term  has degree  $2n-(n-m)=n+m=(2n-(n+m))+2m$ on  both sides  and the
sign on  the LHS is  $(-1)^n$ and on  the RHS it is  $(-1)^{n+m}.$ The
conclusion is that the queried factor is given by
$$\bbox[5px,border:2px solid #00A000]{
Q_{n,m} = (-1)^m \frac{(n+m)!}{(n-m)!}.}$$
