A Requested Explanation

In general: $F(x)=\int_{a}^{f(x)}g(t)dt$. Then $F^{\prime}(x)=g(f(x))\cdot f^{\prime}(x)$.

Or perhaps more generally: $F(x)=\int_{f(x)}^{g(x)}h(t)dt$. Then
to compute the derivate we should first rewrite $F(x)$.

$F(x)=\int_{f(x)}^{a}h(t)dt+\int_{a}^{g(x)}h(t)dt=-\int_{a}^{f(x)}h(t)dt+\int_{a}^{g(x)}h(t)dt$

$F^{\prime}(x)=-h(f(x))\cdot f^{\prime}(x)+h(g(x))\cdot g^{\prime}(x)$.

Thus to apply the above we look at the final exam practice problem 2d. We begin by rewriting $F(x)$ into a form in which we can apply the fundamental theorem of calculus. You should pick a specific $a$, like say $a=0$
\begin{align*}
F(x) & =\int_{\ln x}^{x}\sin(e^{t})dt\\
& =\int_{\ln x}^{a}\sin(e^{t})dt+\int_{a}^{x}\sin(e^{t})dt\\
& =-\int_{a}^{\ln x}\sin(e^{t})dt+\int_{a}^{x}\sin(e^{t})dt\end{align*}

We differentiate, we need to remember the chain rule:
\begin{align*}
F^{\prime}(x) & =-\sin(e^{\ln x})\frac{d}{dx}(\ln x)+\sin(e^{x})\\
& =-\frac{\sin(x)}{x}+\sin(e^{x})\end{align*}

One often sees the chain rule like this:
\begin{align*}
F(x) & =\int_{a}^{u(x)}g(t)dt\\
\frac{d}{dx}F(x) & =g(u(x))\cdot\frac{du}{dx}(x)\end{align*}

Or even more simply:
\begin{align*}
F(x) & =\int_{a}^{u}g(t)dt\\
\frac{dF}{dx} & =g(u)\cdot\frac{du}{dx}\end{align*}