Answer
${\bf{a}}. - \frac{{36}}{{13}};$ ${\bf{b}}. - \frac{{21}}{{10}}$
Work Step by Step
$$\eqalign{
& {\bf{a}}.{\text{ }}{D_x}\left( {g\left[ {f\left( x \right)} \right]} \right){\text{ }} \cr
& {\text{By using the chain rule}} \cr
& {D_x}\left( {g\left[ {f\left( x \right)} \right]} \right) = g'\left[ {f\left( x \right)} \right]f'\left( x \right) \cr
& {\text{At }}x = 2 \cr
& = g'\left[ {f\left( 2 \right)} \right]f'\left( 2 \right) \cr
& {\text{From the table }}f\left( 2 \right) = 4{\text{ and }}f'\left( 2 \right) = - 6 \cr
& g'\left[ {f\left( 2 \right)} \right]f'\left( 2 \right) = g'\left( 4 \right)\left( { - 6} \right) \cr
& {\text{From the table }}g'\left( 4 \right) = \frac{6}{{13}},{\text{ then}} \cr
& g'\left[ {f\left( 2 \right)} \right]f'\left( 2 \right) = \left( {\frac{6}{{13}}} \right)\left( { - 6} \right) \cr
& g'\left[ {f\left( 2 \right)} \right]f'\left( 2 \right) = - \frac{{36}}{{13}} \cr
& \cr
& {\bf{b}}.{\text{ }}{D_x}\left( {g\left[ {f\left( x \right)} \right]} \right){\text{ }} \cr
& {\text{By using the chain rule}} \cr
& {D_x}\left( {g\left[ {f\left( x \right)} \right]} \right) = g'\left[ {f\left( x \right)} \right]f'\left( x \right) \cr
& {\text{At }}x = 3 \cr
& = g'\left[ {f\left( 3 \right)} \right]f'\left( 3 \right) \cr
& {\text{From the table }}f\left( 3 \right) = 2{\text{ and }}f'\left( 3 \right) = - 7 \cr
& g'\left[ {f\left( 3 \right)} \right]f'\left( 3 \right) = g'\left( 2 \right)\left( { - 7} \right) \cr
& {\text{From the table }}g'\left( 2 \right) = \frac{3}{{10}},{\text{ then}} \cr
& g'\left[ {f\left( 3 \right)} \right]f'\left( 3 \right) = \left( {\frac{3}{{10}}} \right)\left( { - 7} \right) \cr
& g'\left[ {f\left( 3 \right)} \right]f'\left( 3 \right) = - \frac{{21}}{{10}} \cr} $$
