Calculate $\frac{d^{100}}{dx^{100}}(\frac{1+x}{\sqrt{1-x}})$

Question

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.

Answer 1

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

$$\frac{100\cdot 198!}{4^{99}\cdot 99!}(1-x)^{-\frac{199}{2}} + \frac{200!}{4^{100}\cdot 100!}(1+x)(1-x)^{-\frac{201}{2}}$$

Answer 2

$$y=\dfrac{1+x}{\sqrt{1-x}}=\dfrac{2-(1-x)}{\sqrt{1-x}}=2(1-x)^{-1/2}-(1-x)^{+1/2}$$

Now use this by writing $$(1-x)^m=(-1)^m(x-1)^m$$

$$\dfrac{d^n(x-1)^m}{dx^n}=m^n(x-1)^{m-n}$$

Answer 3

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 $$ (2n-1)!!=1\cdot3\cdot5\cdots(2n-1). $$