= Solution
A naive absolute bound on the original derivative would be $1/x$, which is not integrable on $[1,\infty)$. Instead make the change of variable $s=tx$, valid for $t>0$:
$$
K(t)=t\int_t^\infty\frac{\cos s}{s^2}\,ds.
$$
The integral is absolutely convergent for every positive lower endpoint. Its integrand is continuous there, so the <fundamental theorem of calculus> gives
$$
\frac d{dt}\int_t^\infty\frac{\cos s}{s^2}\,ds=-\frac{\cos t}{t^2}.
$$
The product rule now yields
$$
\boxed{K'(t)=\int_t^\infty\frac{\cos s}{s^2}\,ds-\frac{\cos t}{t}
=\frac{K(t)-\cos t}{t}.}
$$
This proves <differentiability> and the equation without an unjustified differentiation of a conditionally convergent derivative integral.
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