= Solution
The <Hilbert-transform Fourier multiplier> is bounded, and $\xi^r\widehat\varphi(\xi)$ is integrable for every nonnegative integer $r$. Therefore its inverse <Fourier transform> can be differentiated under the integral arbitrarily many times:
$$
\partial_x^r\mathcal H\varphi(x)=\frac1{2\pi}\int e^{ix\xi}(i\xi)^r[-i\operatorname{sgn}(\xi)]\widehat\varphi(\xi)\,d\xi.
$$
These <derivatives> are continuous by <dominated convergence theorem>. Since $\widehat{\varphi^{(r)}}=(i\xi)^r\widehat\varphi$, \b[the Hilbert transform is smooth and commutes with differentiation]:
$$
\boxed{\mathcal H\varphi\in C^\infty(\mathbb R),\qquad (\mathcal H\varphi)'=\mathcal H(\varphi').}
$$
The Fourier proof avoids differentiating a singular kernel without preserving its <Cauchy principal value> prescription.
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