How to prove $f(x)=\sqrt{x+\sqrt{x+\sqrt{x}}}$ is differentiable?

Question

I've been working on this question for over 2 hours now and I tried to use the limit definition of a derivative to show that it's differentiable but that got me nowhere because I was completely incapable of simplifying the expression I got. If anyone has any leads as to how to start/continue the proof, please explain how I should approach this. Also, I should be able to compute the derivative simply after I prove that it's differentiable, right? (I'm really sorry about this but I've been doing pretty bad in my real analysis course so I'm genuinely just not sure of what I'm doing anymore)

EDIT: What I did was this: 1

I've tried to simplify this by using a few methods, but basically yeah I'm stuck at the first step...

EDIT2: This is the question:

Answer 1

The function is differentiable over $\mathbb{R}_+^*$ as a composition of differentiable functions.

But it is not differentiable at $x=0$ : indeed $$\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} \geq \frac{\sqrt{x}}{x} = \frac{1}{\sqrt{x}}$$ so $$\lim_{x \rightarrow 0} \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} = +\infty$$

Answer 2

To simplify the calculation of the derivative I write $$f(x)=\sqrt{x+\sqrt{x+\sqrt{x}}}=\sqrt{x+g(x)};\;g(x)=\sqrt{x+\sqrt{x}};\;x\ge 0$$ We have $$f'(x) =\frac{g'(x)+1}{2 \sqrt{g(x)+x}}$$ where $$g'(x) =\frac{2 \sqrt{x}+1}{4 \sqrt{x^{3/2}+x^2}}$$ so finally $$f'(x) =\frac{\frac{2 \sqrt{x}+1}{4 \sqrt{x^{3/2}+x^2}}+1}{2 \sqrt{x+\sqrt{x+\sqrt{x}}}}$$