= Solution
Under the usual regularity assumptions, the <Leibniz integral rule> is
$$
\boxed{\frac d{dx}\int_{g(x)}^{f(x)}h(x,t)\,dt=h(x,f(x))f'(x)-h(x,g(x))g'(x)+\int_{g(x)}^{f(x)}h_x(x,t)\,dt.}
$$
There is a genuine domain problem in the following integral as printed in the PDF. Put
$$
q(t)=\frac{2t^4+1}{(t-2)(t^2+3)}.
$$
The point $t=2$ is a simple pole, with
$$
\lim_{t\to2}(t-2)q(t)=\frac{33}{7}\ne0.
$$
For every $x>4$, the lower endpoint $\sin(1/x)$ is less than $2$ and the upper endpoint $\sqrt x$ is greater than $2$. The interval therefore crosses the pole. Since $q(t)=33/[7(t-2)]+O(1)$ near $2$, the two one-sided <improper integrals> diverge logarithmically, with opposite signs. Their divergences do not define an ordinary <integral>. Consequently \b[the ordinary limit in the question is undefined as written]: its underlying integral has no finite value for any $x\geq4$.
A separately specified <Cauchy principal value> does exist for $x>4$. Under that interpretation, <differentiating a principal-value integral across a fixed simple pole> gives
$$
\frac d{dx}\operatorname{PV}\int_{\sin(1/x)}^{\sqrt x}q(t)\,dt=\frac{q(\sqrt x)}{2\sqrt x}+\frac{\cos(1/x)}{x^2}q(\sin(1/x)).
$$
The first term is
$$
\frac{2x^2+1}{2\sqrt x(\sqrt x-2)(x+3)}\longrightarrow1,
$$
and the second tends to zero because $q$ is bounded near $t=0$. Thus the principal-value version has limit $\boxed{1}$. The two conclusions are distinct: this regularised answer requires the additional principal-value convention, which the printed question does not state.
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