= Solution
We prove the required <differentiation under the integral sign> directly. Fix $t_0>0$, and take a compact parameter <interval> $J\subseteq[0,\infty)$ containing $t_0$ in its interior. The jointly continuous <partial derivative> $\partial_t f$ is bounded on the compact rectangle $J\times[0,1]$ by some $M_1$.
For nonzero $h$ sufficiently close to zero, define
$$
D_h(x)=\frac{f(t_0+h,x)-f(t_0,x)}h.
$$
For each $x$, the <mean value theorem> in the parameter gives $|D_h(x)|\leq M_1$. The definition of the <partial derivative> also gives $D_h(x)\to\partial_t f(t_0,x)$. Since $M_1$ is integrable on $[0,1]$, <dominated convergence> applied along any sequence $h_n\to0$ yields
$$
\lim_{h\to0}\frac{F(t_0+h)-F(t_0)}h
=\lim_{h\to0}\int_0^1D_h(x)\,dx
=\int_0^1\partial_t f(t_0,x)\,dx.
$$
Thus
$$
\boxed{F'(t)=\int_0^1\frac{\partial f(t,x)}{\partial t}\,dx\qquad(t>0).}
$$
At $t=0$, repeat the argument on $J=[0,\delta]$ with $h>0$. It proves that the <right derivative> exists and satisfies the same formula at zero. This is the appropriate meaning of differentiability on the closed half-line. In fact $F'$ is continuous there: apply part (i) to the continuous <function> $\partial_t f$.
The proof establishes <differentiation of parameter integrals on compact rectangles> under the exact regularity assumptions used here. In particular, it supplies the theorem's proof instead of assuming an interchange of a <limit> and an <integral>.
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