Answer
$$\left( {\text{a}} \right)2x\sqrt {{x^2} + 1} {\text{ }}\left( {\text{b}} \right) - \frac{1}{{{x^2}}}\sin \left( {\frac{1}{x}} \right)$$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\frac{d}{{dx}}\int_{ - 1}^{{x^2}} {\sqrt {t + 1} dt} \cr
& {\text{Find the derivative using the formula }}\left( {18} \right) \cr
& {\text{Formula }}\left( {{\text{18}}} \right){\text{ }}\frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( t \right)dt} } \right] = f\left( {g\left( x \right)} \right)g'\left( x \right),{\text{ then}} \cr
& \frac{d}{{dx}}\int_{ - 1}^{{x^2}} {\sqrt {t + 1} dt} = \sqrt {{x^2} + 1} \frac{d}{{dx}}\left[ {{x^2}} \right] \cr
& {\text{ }} = 2x\sqrt {{x^2} + 1} \cr
& {\text{Evaluating the integral and then differentiating}} \cr
& \int_{ - 1}^{{x^2}} {\sqrt {t + 1} dt} = \left[ {\frac{2}{3}{{\left( {t + 1} \right)}^{3/2}}} \right]_{ - 1}^{{x^2}} \cr
& {\text{ }} = \frac{2}{3}{\left( {{x^2} + 1} \right)^{3/2}} - \frac{2}{3}{\left( 0 \right)^{3/2}} \cr
& {\text{ }} = \frac{2}{3}{\left( {{x^2} + 1} \right)^{3/2}} \cr
& \frac{d}{{dx}}\left[ {\frac{2}{3}{{\left( {{x^2} + 1} \right)}^{3/2}}} \right] = \frac{2}{3}\left( {\frac{3}{2}} \right){\left( {{x^2} + 1} \right)^{1/2}}\left( {2x} \right) \cr
& {\text{ }} = 2x\sqrt {{x^2} + 1} \cr
& \cr
& \left( {\text{b}} \right)\frac{d}{{dx}}\int_\pi ^{1/x} {\sin tdt} \cr
& {\text{Find the derivative using the formula }}\left( {18} \right) \cr
& \frac{d}{{dx}}\int_\pi ^{1/x} {\sin tdt} = \sin \left( {\frac{1}{x}} \right)\frac{d}{{dx}}\left[ {\frac{1}{x}} \right] \cr
& {\text{ }} = - \frac{1}{{{x^2}}}\sin \left( {\frac{1}{x}} \right) \cr
& {\text{Evaluating the integral and then differentiating}} \cr
& \int_\pi ^{1/x} {\sin tdt} = \left[ { - \cos t} \right]_\pi ^{1/x} \cr
& {\text{ }} = - \cos \left( {\frac{1}{x}} \right) + \cos \pi \cr
& \frac{d}{{dx}}\left[ { - \cos \left( {\frac{1}{x}} \right) + \cos \pi } \right] = \sin \left( {\frac{1}{x}} \right)\left( { - \frac{1}{{{x^2}}}} \right) + 0 \cr
& {\text{ }} = - \frac{1}{{{x^2}}}\sin \left( {\frac{1}{x}} \right) \cr} $$
