<s>[INST] Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$ Calculate $$\frac{d^{100}}{dx^{100}}\left(\frac{1+x}{\sqrt{1-x}}\right).$$
I gathered that I can use Leibniz's formula, so the differentiation can be represented by the following sum:
$$ \sum^{100}_{r=0} \binom{100}{r}\left[\frac{d^{100-r}}{dx^{100-r}}(1+x)\right]\left[ \frac{d^r}{dx^r}(1-x)^{-\frac{1}{2}} \right].$$
Since we know that $ \frac{d^{2}}{dx^2}(1+x) = 0 $, we can simplify the above sum to the following:
$$ 0+\ldots+\binom{100}{98}\left[\frac{d^{2}}{dx^2}(1+x)\right] \left[ \frac{d^{98}}{dx^{98}}(1-x)^{-\frac{1}{2}} \right] 
+\binom{100}{99}\left[\frac{d}{dx}(1+x)\right] \left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] +\binom{100}{100}(1+x) \left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right], $$
which then gives us,
$$ 100\left[ \frac{d^{99}}{dx^{99}}(1-x)^{-\frac{1}{2}} \right] + (1+x)\left[ \frac{d^{100}}{dx^{100}}(1-x)^{-\frac{1}{2}} \right].$$
This is where I am stucked. I'm not sure resolve those differentials.
 [/INST] Note
$$\frac{1}{\sqrt{1-x}}=\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}$$
and hence
$$\frac{1+x}{\sqrt{1-x}}=(1+x)\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}=\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^n}{2^n n!}+\sum_{n = 0}^{\infty}\frac{(2n - 1)!!x^{n+1}}{2^n n!}.$$
So
$$ \frac{d^{100}}{dx^{100}}\frac{1+x}{\sqrt{1-x}}=\cdots. $$
Here
$$ (2n-1)!!=1\cdot3\cdot5\cdots(2n-1). $$
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