Answer
$$\left( {\text{a}} \right) - \cos \left( {{x^3}} \right){\text{ }}\left( {\text{b}} \right) - {\tan ^2}x$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\frac{d}{{dx}}\int_x^\pi {\cos \left( {{t^3}} \right)} dt \cr
& {\text{Differentiate, use }}\frac{d}{{dx}}\left[ {\int_x^a {f\left( t \right)dt} } \right] = - f\left( x \right),{\text{ then}} \cr
& \frac{d}{{dx}}\left[ {\int_x^\pi {\cos \left( {{t^3}} \right)} dt} \right] = - \cos \left( {{x^3}} \right) \cr
& \cr
& \left( {\text{b}} \right)\frac{d}{{dx}}\int_{\tan x}^3 {\frac{{{t^2}}}{{1 + {t^2}}}} dt \cr
& {\text{Differentiate, use }}\frac{d}{{dx}}\left[ {\int_{g\left( x \right)}^a {f\left( t \right)dt} } \right] = - f\left( {g\left( x \right)} \right)g'\left( x \right),{\text{ then}} \cr
& \frac{d}{{dx}}\left[ {\int_{\tan x}^3 {\frac{{{t^2}}}{{1 + {t^2}}}} dt} \right] = - \frac{{{{\left( {\tan x} \right)}^2}}}{{1 + {{\left( {\tan x} \right)}^2}}}\frac{d}{{dx}}\left[ {\tan x} \right] \cr
& {\text{ }} = - \frac{{{{\tan }^2}x}}{{1 + {{\tan }^2}x}}\left( {{{\sec }^2}x} \right) \cr
& {\text{ }} = - \frac{{{{\tan }^2}x}}{{{{\sec }^2}x}}\left( {{{\sec }^2}x} \right) \cr
& {\text{ }} = - {\tan ^2}x \cr} $$
