Answer
${\bf{a}}. - \frac{3}{2}$; ${\bf{b}}. - \frac{{24}}{{11}}$
Work Step by Step
$$\eqalign{
& {\bf{a}}.{\text{ }}{D_x}\left( {f\left[ {g\left( x \right)} \right]} \right){\text{ }} \cr
& {\text{By using the chain rule}} \cr
& {D_x}\left( {f\left[ {g\left( x \right)} \right]} \right) = f'\left[ {g\left( x \right)} \right]g'\left( x \right) \cr
& {\text{At }}x = 2 \cr
& = f'\left[ {g\left( 2 \right)} \right]g'\left( 2 \right) \cr
& {\text{From the table }}g\left( 2 \right) = 1{\text{ and }}g'\left( 2 \right) = \frac{3}{{10}} \cr
& f'\left[ {g\left( 2 \right)} \right]g'\left( 2 \right) = f'\left( 1 \right)\left( {\frac{3}{{10}}} \right) \cr
& {\text{From the table }}f'\left( 1 \right) = - 5,{\text{ then}} \cr
& f'\left[ {g\left( 2 \right)} \right]g'\left( 2 \right) = \left( { - 5} \right)\left( {\frac{3}{{10}}} \right) \cr
& f'\left[ {g\left( 2 \right)} \right]g'\left( 2 \right) = - \frac{3}{2} \cr
& \cr
& {\bf{b}}.{\text{ }}{D_x}\left( {f\left[ {g\left( x \right)} \right]} \right){\text{ }} \cr
& {\text{By using the chain rule}} \cr
& {D_x}\left( {f\left[ {g\left( x \right)} \right]} \right) = f'\left[ {g\left( x \right)} \right]g'\left( x \right) \cr
& {\text{At }}x = 3 \cr
& = f'\left[ {g\left( 3 \right)} \right]g'\left( 3 \right) \cr
& {\text{From the table }}g\left( 3 \right) = 2{\text{ and }}g'\left( 3 \right) = \frac{4}{{11}} \cr
& f'\left[ {g\left( 3 \right)} \right]g'\left( 3 \right) = f'\left( 2 \right)\left( {\frac{4}{{11}}} \right) \cr
& {\text{From the table }}f'\left( 2 \right) = - 6,{\text{ then}} \cr
& f'\left[ {g\left( 3 \right)} \right]g'\left( 3 \right) = \left( { - 6} \right)\left( {\frac{4}{{11}}} \right) \cr
& f'\left[ {g\left( 3 \right)} \right]g'\left( 3 \right) = - \frac{{24}}{{11}} \cr} $$
