We are to differentiate

$$f ( x ) = \int _ { 0 } ^ { 2 x } \left( t ^ { 2 } - x ^ { 2 } \right) \sin 3 t \, d t$$

with respect to $x$.

(1) We know that if $g ( t )$ is a continuous function and $G ( t )$ is one of its primitive functions, then

$$\int _ { 0 } ^ { 2 x } g ( t ) d t = G ( 2 x ) - G ( 0 )$$

By differentiating both sides of this equality with respect to $x$, we have

$$\frac { d } { d x } \int _ { 0 } ^ { 2 x } g ( t ) d t = \mathbf { A }$$

where $\mathbf{A}$ is the appropriate expression from among the following (0) $\sim$ (7).

(0) $g ( x )$\\
(1) $\frac { 1 } { 2 } g ( x )$\\
(2) $2 g ( x )$\\
(3) $g ( 2 x )$\\
(4) $\frac { 1 } { 2 } g ( 2 x )$\\
(5) $2 g ( 2 x )$\\
(6) $g ( x ) - g ( 0 )$\\
(7) $g ( 2 x ) - g ( 0 )$

(2) We know that $f ( x ) = \int _ { 0 } ^ { 2 x } t ^ { 2 } \sin 3 t \, d t - \int _ { 0 } ^ { 2 x } x ^ { 2 } \sin 3 t \, d t$.

Since

$$\frac { d } { d x } \int _ { 0 } ^ { 2 x } t ^ { 2 } \sin 3 t \, d t = \mathbf { B } x ^ { 2 } \sin \mathbf { C } x$$

and

$$\frac { d } { d x } \int _ { 0 } ^ { 2 x } x ^ { 2 } \sin 3 t \, d t = \frac { \mathbf { D } } { \mathbf { E } } x ( - \cos \mathbf { F } x + \mathbf { G } + \mathbf { H } x \sin \mathbf { I } x )$$

we obtain

$$f ^ { \prime } ( x ) = \frac { \mathbf { D } } { \mathbf { E } } x ( \cos \mathbf { J } x - \mathbf { K } + \mathbf { L } x \sin \mathbf { M } x )$$