If $f ( x ) = \int _ { 1 } ^ { x ^ { 3 } } \frac { 1 } { 1 + \ln t } d t$ for $x \geq 1$, then $f ^ { \prime } ( 2 ) =$
(A) $\frac { 1 } { 1 + \ln 2 }$
(B) $\frac { 12 } { 1 + \ln 2 }$
(C) $\frac { 1 } { 1 + \ln 8 }$
(D) $\frac { 12 } { 1 + \ln 8 }$

Let $f , g$ be continuous functions from $[ 0 , \infty )$ to itself, $$h ( x ) = \int _ { 2 ^ { x } } ^ { 3 ^ { x } } f ( t ) d t , x > 0$$ and $$F ( x ) = \int _ { 0 } ^ { h ( x ) } g ( t ) d t , x > 0$$ If $F ^ { \prime }$ is the derivative of $F$, then for $x > 0$,
(A) $F ^ { \prime } ( x ) = g ( h ( x ) )$.
(B) $F ^ { \prime } ( x ) = g ( h ( x ) ) \left[ f \left( 3 ^ { x } \right) - f \left( 2 ^ { x } \right) \right]$.
(C) $F ^ { \prime } ( x ) = g ( h ( x ) ) \left[ x 3 ^ { x - 1 } f \left( 3 ^ { x } \right) - x 2 ^ { x - 1 } f \left( 2 ^ { x } \right) \right]$.
(D) $F ^ { \prime } ( x ) = g ( h ( x ) ) \left[ 3 ^ { x } f \left( 3 ^ { x } \right) \ln 3 - 2 ^ { x } f \left( 2 ^ { x } \right) \ln 2 \right]$.

Let $f$ be a non-negative function in $[ 0,1 ]$ and twice differentiable in $( 0,1 )$. If $\int _ { 0 } ^ { x } \sqrt { 1 - \left( f ^ { \prime } ( t ) \right) ^ { 2 } } \mathrm { dt } = \int _ { 0 } ^ { x } f ( \mathrm { t } ) \mathrm { dt } , 0 \leq x \leq 1$ and $f ( 0 ) = 0$, then $\lim _ { x \rightarrow 0 } \frac { 1 } { x ^ { 2 } } \int _ { 0 } ^ { x } f ( \mathrm { t } ) \mathrm { dt } :$
(1) does not exist
(2) equals 0
(3) equals 1
(4) equals $\frac { 1 } { 2 }$

Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f\left(\frac{\pi}{4}\right) = \sqrt{2}$, $f\left(\frac{\pi}{2}\right) = 0$ and $f'\left(\frac{\pi}{2}\right) = 1$ and let $g(x) = \int_x^{\pi/4} (f'(t)\sec t + \tan t \cdot f(t)\sec t)\,dt$. Then $\lim_{x \to \pi/2} \frac{g(x)}{(x - \pi/2)^2}$ is equal to $\_\_\_\_$.

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