Answer
$$\left( {\text{a}} \right) - \frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}}{\text{ }}\left( {\text{b}} \right)\frac{1}{{{x^2}}}{\cos ^3}\left( {\frac{1}{x}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\frac{d}{{dx}}\int_x^0 {\frac{1}{{{{\left( {{t^2} + 1} \right)}^2}}}} dt \cr
& {\text{Differentiate, use }}\frac{d}{{dx}}\left[ {\int_x^a {f\left( t \right)dt} } \right] = - f\left( x \right),{\text{ then}} \cr
& \frac{d}{{dx}}\left[ {\int_x^0 {\frac{1}{{{{\left( {{t^2} + 1} \right)}^2}}}} dt} \right] = - \frac{1}{{{{\left( {{x^2} + 1} \right)}^2}}} \cr
& \cr
& \left( {\text{b}} \right)\frac{d}{{dx}}\int_{1/x}^\pi {{{\cos }^3}t} dt \cr
& {\text{Differentiate, use }}\frac{d}{{dx}}\left[ {\int_{g\left( x \right)}^a {f\left( t \right)dt} } \right] = - f\left( {g\left( x \right)} \right)g'\left( x \right),{\text{ then}} \cr
& \frac{d}{{dx}}\left[ {\int_{1/x}^\pi {{{\cos }^3}t} dt} \right] = - {\cos ^3}\left( {\frac{1}{x}} \right)\frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr
& {\text{ }} = - {\cos ^3}\left( {\frac{1}{x}} \right)\left( { - \frac{1}{{{x^2}}}} \right) \cr
& {\text{ }} = \frac{1}{{{x^2}}}{\cos ^3}\left( {\frac{1}{x}} \right) \cr} $$
