<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] The derivative of $\sqrt{1-x}$ follows a pattern
$$f'(x) = \frac{1}{2}(1-x)^{-\frac{3}{2}}$$
$$f''(x) = \frac{1\cdot 3}{2\cdot 2}(1-x)^{-\frac{5}{2}}$$
$$\vdots$$
$$f^{(n)}(x) = \frac{(2n)!}{4^n n!}(1-x)^{-\frac{(2n+1)}{2}}$$
a product of odds divided by a power of $2$ which makes your derivative 
$$\frac{100\cdot 198!}{4^{99}\cdot 99!}(1-x)^{-\frac{199}{2}} + \frac{200!}{4^{100}\cdot 100!}(1+x)(1-x)^{-\frac{201}{2}}$$
</s>