Answer
the derivative of the function
$$
y=\int_{\sqrt{x}}^{x} \frac{\cos \theta}{\theta} d \theta
$$
is
$$
y^{\prime} =\frac{2 \cos x-\cos \sqrt{x}}{2 x}
$$
Work Step by Step
$$
y=\int_{\sqrt{x}}^{x} \frac{\cos \theta}{\theta} d \theta
$$
we must split this integral into two parts as follows :
$$
\begin{aligned}
y=\int_{\sqrt{x}}^{x} \frac{\cos \theta}{\theta} d \theta &=\int_{1}^{x} \frac{\cos \theta}{\theta} d \theta+\int_{\sqrt{x}}^{1} \frac{\cos \theta}{\theta} d \theta \\
&=\int_{1}^{x} \frac{\cos \theta}{\theta} d \theta-\int_{1}^{\sqrt{x}} \frac{\cos \theta}{\theta} d \theta
\end{aligned}
$$
the derivative of the function $y$ is given by the following:
$$
\begin{aligned}
y^{\prime} &=\frac{\cos x}{x} . \frac{d}{dx} (x)-\frac{\cos \sqrt{x}}{\sqrt{x}}\frac{d}{dx} (\sqrt x)\\
&=\frac{\cos x}{x} . (1)-\frac{\cos \sqrt{x}}{\sqrt{x}} . \frac{1}{2 \sqrt{x}} \\
&=\frac{2 \cos x-\cos \sqrt{x}}{2 x}
\end{aligned}
$$
