= Solution
A direct <derivative> bound by $1/x$ would not be <integrable>, so the earlier theorem cannot be applied that way. Instead, for $t>0$ make the legitimate substitution $s=tx$ in the <absolutely convergent> integral:
$$
K(t)=t\int_t^\infty\frac{\cos s}{s^2}\,ds.
$$
The tail integral has <continuous> integrand away from zero and converges absolutely. The <fundamental theorem of calculus> therefore gives <derivative> $-\cos t/t^2$ with respect to its lower endpoint. The <product rule> yields
$$
\boxed{K'(t)=\int_t^\infty\frac{\cos s}{s^2}\,ds-\frac{\cos t}t=\frac{K(t)-\cos t}t.}
$$
This establishes <differentiability> carefully without an invalid domination argument.
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