Answer
$$\left( {\text{a}} \right)\frac{3}{x}{\text{ }}\left( {\text{b}} \right)1$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\frac{d}{{dx}}\int_1^{{x^3}} {\frac{1}{t}dt} \cr
& {\text{Find the derivative using the formula }}\left( {18} \right) \cr
& {\text{Formula }}\left( {{\text{18}}} \right){\text{ }}\frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( t \right)dt} } \right] = f\left( {g\left( x \right)} \right)g'\left( x \right),{\text{ then}} \cr
& \frac{d}{{dx}}\int_1^{{x^3}} {\frac{1}{t}dt} = \frac{1}{{{x^3}}}\frac{d}{{dx}}\left[ {{x^3}} \right] \cr
& {\text{ }} = \frac{1}{{{x^3}}}\left( {3{x^2}} \right) \cr
& {\text{ }} = \frac{3}{x} \cr
& {\text{Evaluating the integral and then differentiating}} \cr
& \int_1^{{x^3}} {\frac{1}{t}dt} = \ln {x^3} - \ln 1 \cr
& {\text{ }} = \ln {x^3} \cr
& \frac{d}{{dx}}\left[ {\ln {x^3}} \right] = \frac{{3{x^2}}}{{{x^3}}} \cr
& {\text{ }} = \frac{3}{x} \cr
& \cr
& \left( {\text{b}} \right)\frac{d}{{dx}}\int_1^{\ln x} {{e^t}dt} \cr
& {\text{Find the derivative using the formula }}\left( {18} \right) \cr
& \frac{d}{{dx}}\int_1^{\ln x} {{e^t}dt} = {e^{\ln x}}\frac{d}{{dx}}\left[ {\ln x} \right] \cr
& {\text{ }} = x\left( {\frac{1}{x}} \right) \cr
& {\text{ }} = 1 \cr
& {\text{Evaluating the integral and then differentiating}} \cr
& \int_1^{\ln x} {{e^t}dt} = {e^{\ln x}} - {e^1} \cr
& {\text{ }} = x - 1 \cr
& \frac{d}{{dx}}\left[ {x - 1} \right] = 1 \cr} $$
