Answer
$$
y=\int_{\sqrt{x}}^{x} \frac{e^{t}}{t} d t
$$
The derivative of the given function is
$$
\frac{d y}{d x}=-\frac{e^{\sqrt{x}}}{2 x} +\frac{e^{x}}{x}=\frac{2e^{x}-e^{\sqrt{x}}}{2x}
$$
Work Step by Step
$$
y=\int_{\sqrt{x}}^{x} \frac{e^{t}}{t} d t
$$
$$
\begin{split}
y &=\int_{\sqrt{x}}^{x} \frac{e^{t}}{t} d t \\
& =\int_{\sqrt{x}}^{1} \frac{e^{t}}{t} d t+\int_{1}^{x} \frac{e^{t}}{t} d t \\
& =-\int_{1}^{\sqrt{x}} \frac{e^{t}}{t} d t+\int_{1}^{x} \frac{e^{t}}{t} d t
\end{split}
$$
$ \Rightarrow $
$$
\frac{d y}{d x}=-\frac{d}{d x}\left(\int_{1}^{\sqrt{x}} \frac{e^{t}}{t} d t\right)+\frac{d}{d x}\left(\int_{1}^{x} \frac{e^{t}}{t} d t\right)
$$
Let $ u = \sqrt {x}$ . Then
$$
\begin{split}
\frac{d}{d x} \int_{1}^{\sqrt{x}} \frac{e^{t}}{t} d t&=\frac{d}{d x} \int_{1}^{u} \frac{e^{t}}{t} d t \\
&=\frac{d}{d u}\left(\int_{1}^{u} \frac{e^{t}}{t} d t\right) \frac{d u}{d x} \\
&=\frac{e^{u}}{u} \cdot \frac{1}{2 \sqrt{x}} \\
&=\frac{e^{\sqrt{x}}}{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}} \\
&=\frac{e^{\sqrt{x}}}{2 x}
\end{split}
$$
so, the derivative of the given function is
$$
\frac{d y}{d x}=-\frac{e^{\sqrt{x}}}{2 x} +\frac{e^{x}}{x}=\frac{2e^{x}-e^{\sqrt{x}}}{2x}
$$
