Positive integer powers of a real valued function on real line that has an anti-derivative

Question

Let $f : \mathbb R \to \mathbb R$ be a differentiable function.

For every integer $n \ge 2$, does there exist a differentiable function $h_n : \mathbb R \to \mathbb R$ such that $h_n'(x)= (f'(x))^n, \forall x \in \mathbb R$ ?

Answer 1

Note that $$g(x)=\begin{cases}\sin\frac1x&x\ne 0\\0&x=0\end{cases}$$ and $$h(x)=\begin{cases}\cos\frac1x&x\ne 0\\0&x=0\end{cases}$$ are derivatives. If the claim were true, so would be $g^2$ and $h^2$ and their sum $$g(x)^2+h(x)^2=\begin{cases}1&x\ne 0\\0&x=0\end{cases} $$

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For completeness' sake: How do we see that $g$ and $h$ are derivatives?

Consider the functions $$ k_h(x)=\begin{cases}x^2\sin\frac1x&x\ne 0\\0 &x=0\end{cases}\\ k_g(x)=\begin{cases}x^2\cos\frac1x&x\ne 0\\0 &x=0\end{cases}$$ Then one verifies $$k_h'(x)=\begin{cases}2x\sin \frac 1x -\cos\frac1x&x\ne 0\\0&x=0\end{cases}\\ k_g'(x)=\begin{cases}2x\cos \frac 1x+\sin\frac1x&x\ne 0\\0&x=0\end{cases}$$ so that $g$ differs from $k_g'$ and $h$ differs from $-k_h'$ by a continuous function. Continuous functions are derivatives by the Fundamental Theorem, sums of derivatives are derivatives, hence $g$ and $h$ are derivatives, as desired.