Answer
$$
y=\int_{2 x}^{3 x+1} \sin \left(t^{4}\right) d t
$$
The derivative of the given function is:
$$
y^{\prime} =3 \sin \left[(3 x+1)^{4}\right]-2 \sin \left[(2 x)^{4}\right].
$$
Work Step by Step
$$
y=\int_{2 x}^{3 x+1} \sin \left(t^{4}\right) d t
$$
$$
\begin{split}
y &=\int_{2 x}^{3 x+1} \sin \left(t^{4}\right) d t \\
&=\int_{2 x}^{0} \sin \left(t^{4}\right) d t+\int_{0}^{3 x+1} \sin \left(t^{4}\right) d t \\
&=\int_{0}^{3 x+1} \sin \left(t^{4}\right) d t-\int_{0}^{2 x} \sin \left(t^{4}\right) d t
\end{split}
$$
$\Rightarrow $
$$
\begin{split}
y^{\prime} &=\sin \left[(3 x+1)^{4}\right] \cdot \frac{d}{d x}(3 x+1)-\sin \left[(2 x)^{4}\right] \cdot \frac{d}{d x}(2 x) \\
&=3 \sin \left[(3 x+1)^{4}\right]-2 \sin \left[(2 x)^{4}\right]
\end{split}
$$
So, the derivative of the given function is:
$$
y^{\prime} =3 \sin \left[(3 x+1)^{4}\right]-2 \sin \left[(2 x)^{4}\right].
$$
