How to proceed for this problem on analysis Find all differentiable functions $f\colon [0,\infty)\to [0,\infty)$ for which $f(0)=0$ and $f^{\prime}(x^2)=f(x)$ for any $x\in [0,\infty)$. 

I have tried to reduce to the form $f(x)=f'(x^{\frac{1}{2^n}})$. but it is not coming. Is there any other way?
First note that $f$ and $f'$ are both nondecreasing.

By MVT, there is a $0 &lt; \lambda &lt; 1$ such that $f(1)=f'(\lambda)$. But $f'(\lambda)=f(\sqrt{ \lambda })$ and since $f$ is nondecreasing, it should be constant on $[\sqrt{ \lambda }, 1]$. Therefore, if $\sqrt{ \lambda } &lt; x &lt; 1$, we have $f'(x)=0=f(\sqrt{x})$. By continuity, $f(1)=0$.

Take $0 &lt; \delta &lt; 1$ and suppose that $f(n)=0$. Then
$$f(n + \delta) = \int_n^{n + \delta} f'(x)dx $$
$$ \le \delta f'(n + \delta) $$
$$ = \delta f(\sqrt{n + \delta}) $$
$$ \le \delta f(n + \delta) $$ 

The last inequality follows because for $n \ge 1$, we have $1 &lt; \sqrt{n + \delta} &lt; n + \delta$. So $f(n+\delta) \le 0$ and then $f(n + \delta)=0$. By continuity, $f(n+1)=0$ and the induction is complete.
We prove this in two steps. First, I prove $f$ must be identically $0$ on $[0,1]$, then I prove the result for $[0, \infty)$. We prove the more general result where $f$ is assumed $f:[0, \infty) \to \mathbb{R}$


  $f \equiv 0$ on $[0,1]$. 


Proceeding by contradiction, suppose $f$ is not identically $0$ on $[0,1]$. By the extreme value theorem, $|f|$ attains a maximum value on $[0,1]$. Let $c$ be an arbitrary maximizer of $|f|$ on $[0,1]$ and note $|f(c)|&gt;0$. Note that this would also imply that $c^2$ is a maximizer of $|f'|$ on $[0,1]$, since $|f(c)| = |f'(c^2)|$. 

We have that $$|f(c)| = \left|\int_{0}^{c} f'(t) \ dt \right|\leq \int_{0}^{c} |f'(t)| dt \leq \int_{0}^{c} |f'(c^2)| dt = c|f'(c^2)| = c|f(c)|$$  In particular, $$|f(c)| \leq c|f(c)|$$

Since $|f(c)|&gt;0$, we can divide through by $|f(c)|$ to simplify the inequality to $$1 \leq c$$

The above argument implies that any maximizer $c$ of $f$ on $[0,1]$ must satisfy $1 \leq c$. Since $c \in [0,1]$, this implies that $c=1$ is the unique maximizer of $f$ on $[0,1]$. 

Note by the mean value theorem $\frac{f(1) - f(0)}{1-0} = f(1) = f'(\delta)$ for some $\delta \in (0,1)$. Hence $f(1) = f'(\delta) = f(\sqrt{\delta})$ and $\sqrt{\delta} \neq 1$ since $\delta \neq 1$. But this means that $\sqrt{\delta}$ is a maximizer of $|f|$ on $[0,1]$ not equal to $1$, which contradicts that $c=1$ is the unique maximizer. 


  $f \equiv 0$ on $[0, \infty)$. 


By contradiction. Suppose $f$ satisfies the given conditions and yet is not identically $0$. Put $n_0 = \text{max}\{n \in \mathbb{N}: f \ \text{is identically zero on} \ [0,n]\}$. We have that $f(x) \neq 0$ for some $ x \in [n_0, n_0+1]$. Note that $n_0 \geq 1$ since we've already done $[0,1]$. Choose $c$ which maximizes $|f|$ on $[n_0, n_0 + 1]$ .  We can assume without loss of generality that $f(c) &gt; 0$, since otherwise define $g=-f$, noting that $g$ satisfies the hypotheses for the problem, and use that to obtain the contradiction instead. Since $f$ is identically $0$ on $[0, n_0]$, it actually follows that $f(c) = \text{max}_{x \in [0, n_0+1]} f(x)$

We have that $f(c)  = f'(\delta)(c-n_0) = f(\sqrt{\delta})(c-n_0)$ where $\delta \in (n_0, c)$ by the mean value theorem. Since $c-n_0 \leq 1$, this implies $f(\sqrt{\delta}) \geq f(c)$. Certainly it's impossible that $f(\sqrt{\delta}) &gt; f(c)$, since $\sqrt{\delta} \in [0, n_0 + 1]$ and $c$ maximizes $f$ on $[0, n_0 + 1]$. Hence we have that $f(\sqrt{\delta}) = f(c)$, so, in fact, $\sqrt{\delta}$ also maximizes $f$ on $[0, n_0 + 1]$. 

While it should be clear that $\sqrt{\delta} \in [0, n_0 + 1]$, one may also note more precisely that we must have $\sqrt{\delta} \in (n_0, n_0 + 1]$. This is because $f$ is identically $0$ in $[0, n_0]$ but $f(\sqrt{\delta})&gt;0$. Moreover, note $\sqrt{\delta} &lt; \delta$ since $\delta&gt;n_0 \geq 1$. With this in mind, we have that $n_0 &lt; \sqrt{\delta} &lt; \delta &lt; n_0 + 1$, so in particular $\sqrt{\delta} - n_0 &lt; 1$. 

We use the mean value theorem one more time to finish off the proof. Note that $f(\sqrt{\delta}) - f(n_0) = f(\sqrt{\delta}) = f'(\gamma)(\sqrt{\delta} - n_0) = f(\sqrt{\gamma})(\sqrt{\delta} - n_0)$ where $\gamma \in (n_0, \sqrt{\delta})$. Since $\sqrt{\delta} - n_0&lt;1$, we may conclude that $f(\sqrt{\gamma}) &gt; f(\sqrt{\delta})$, which contradicts that $\sqrt{\delta}$ maximizes $f$ on $[0,n_0 + 1]$. QED.