How can I find the $n$ th derivatve of this function

Question

Let $f(x)=\sqrt{1+\sqrt{1-x}}\,\,\,\,\,\ \forall x\in[0,1].$

It is not difficult to find first few derivatives of this function as

$$f'(x)=-\dfrac1{2^2}(1+\sqrt{1-x})^{-1/2}(1-x)^{-1/2}$$

$$f''(x)=-\frac1{2^4}(1+\sqrt{1-x})^{-3/2}(1-x)^{-1}-\frac1{2^3}(1+\sqrt{1-x})^{-1/2}(1-x)^{-3/2}$$

$$f'''(x)=-\frac3{2^6}(1+\sqrt{1-x})^{-5/2}(1-x)^{-3/2}-\frac3{2^5}(1+\sqrt{1-x})^{-3/2}(1-x)^{-2}-\frac3{2^4}(1+\sqrt{1-x})^{-1/2}(1-x)^{-5/2}$$

However it is difficult to observe a pattern between these derivatives.
How can I find the $n$th derivative of $f$?

Answer 1

Edit. One may use the following result.

Hint. Let $f$ be $n$-times differentiable over some interval $(0,b).$ Then

$$ \left(\frac{d}{dx}\right)^n\left[f(\sqrt{x})\frac{}{}\right]=\sum_{k=0}^{n-1}(-1)^k\frac{(n+k-1)!}{k!(n-k-1)!}\frac{1}{(2\sqrt{x})^{n+k}}f^{(n-k)}(\sqrt{x}).\tag1 $$

Since $$ \sqrt{1+\sqrt{1-x}}=\frac1{\sqrt{2}}\left(\sqrt{1+\sqrt{x}}+\sqrt{1-\sqrt{x}}\right) $$then one may apply $(1)$ to $$ f(x)=\frac1{\sqrt{2}}\left(\sqrt{1+x}+\sqrt{1-x}\right),\qquad 0<x<1,\tag2 $$ observing that $$ f^{(k)}(x)=\frac1{\sqrt{2}}\left(\frac{(-1)^{k-1}(2k)!}{(2k-1)2^{2k}k!}\frac1{(1+x)^{k-\frac12}}-\frac{(2k)!}{(2k-1)2^{2k}k!}\frac1{(1-x)^{k-\frac12}}\right). \tag3 $$

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Answer 2

Just a partial answer. Since $$ f(x)=\frac{1}{\sqrt{2}}\left(\sqrt{1+\sqrt{x}}+\sqrt{1-\sqrt{x}}\right)\tag{1}$$ and $$ \sqrt{1-x} = -\sum_{n\geq 0}\frac{\binom{2n}{n}}{(2n-1)4^n}x^n \tag{2}$$ we have that: $$ \sqrt{1-x}+\sqrt{1+x} = -2\sum_{n\geq 0}\frac{\binom{4n}{2n}}{(4n-1)16^n} x^{2n} $$ and: $$ f(x) = -\sqrt{2}\sum_{n\geq 0}\frac{\binom{4n}{2n}}{(4n-1)16^n}\,x^{n} \tag{3} $$ so that: $$ f^{(n)}(0) = -\sqrt{2}\,\frac{\binom{4n-2}{2n-1}(n-1)!}{16^n}\tag{4} $$ for any $n\geq 1$. About the general expression of $f^{(n)}(x)$, there is no shame in exploiting our pattern matching abilities. It looks like: $$ f^{(n)}(x)=\sum_{k=1}^{n} c(n,k) \left(1+\sqrt{1-x}\right)^{-(2k-1)/2}(1-x)^{n-k/2} \tag{5}$$ and it is enough to differentiate both sides of $(5)$ wrt to $x$ to derive a recurrence relation for the coefficients $c(n,k)$.