= Solution
Fix any real $t_0$ and restrict $t$ to $|t-t_0|<1$. The parameter derivative is
$$
\partial_tu(x,t)=\frac{e^{tx}}{\sqrt{1-x}},\qquad
|\partial_tu(x,t)|\leq\frac{e^{|t_0|+1}}{\sqrt{1-x}}.
$$
This bound is integrable on $(0,1)$ and is uniform throughout the parameter neighborhood. Part (ii) supplied integrability at $t_0$, so <dominated differentiation under the integral sign> proves differentiability there. As $t_0$ was arbitrary,
$$
\boxed{F'(t)=\int_0^1\frac{e^{tx}}{\sqrt{1-x}}\,dx\quad(t\in\mathbb R).}
$$
For $t>0$, $e^{tx}\geq1$; for $t<0$, $e^{tx}\leq1$. Since $\int_0^1(1-x)^{-1/2}\,dx=2$, these comparisons give
$$
\boxed{F'(t)\geq2\ (t>0),\qquad F'(t)\leq2\ (t<0),\qquad F'(0)=2.}
$$
The first two inequalities are actually strict, because the exponential differs strictly from one throughout $(0,1)$. Differentiating once more under the same type of bound gives $F''(t)=\int_0^1xe^{tx}/\sqrt{1-x}\,dx>0$. Hence $F$ is <strictly convex> and lies above its tangent $2t$ at zero. These properties are collected in <exponential integral with a square-root endpoint singularity>.
\Image[/past-exam-of-the-integration-course-of-the-university-of-oxford/2017-exponential-endpoint-integral.png]
{title=The parameter integral is convex, and its derivative crosses two at zero}
{description=The endpoint singularity is removed numerically using $x=1-s^2$. The left panel compares $F$ with its tangent $2t$; the right panel displays the derivative inequalities proved in the solution.}
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