Answer
$$\eqalign{
& \left( {\text{a}} \right){\text{Proved}} \cr
& \left( {\text{b}} \right){\text{Proved}} \cr} $$
Work Step by Step
$$\eqalign{
& \left( {\text{a}} \right)\frac{d}{{dx}}\left[ {\int_x^a {f\left( t \right)dt} } \right] = - f\left( x \right) \cr
& {\text{Apply the property }}\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx \cr
& \frac{d}{{dx}}\left[ {\int_x^a {f\left( t \right)dt} } \right] = \frac{d}{{dx}}\left[ { - \int_a^x {f\left( t \right)dt} } \right] = - \frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] \cr
& {\text{Where }}\frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = f\left( x \right),{\text{ then}} \cr
& - \frac{d}{{dx}}\left[ {\int_a^x {f\left( t \right)dt} } \right] = - f\left( x \right) \cr
& \cr
& \left( {\text{b}} \right)\frac{d}{{dx}}\left[ {\int_{g\left( x \right)}^a {f\left( t \right)dt} } \right] = - f\left( {g\left( x \right)} \right)g'\left( x \right) \cr
& {\text{Apply the property }}\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx \cr
& \frac{d}{{dx}}\left[ {\int_{g\left( x \right)}^a {f\left( t \right)dt} } \right] = \frac{d}{{dx}}\left[ { - \int_a^{g\left( x \right)} {f\left( t \right)dt} } \right] \cr
& = - \frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( t \right)dt} } \right] \cr
& {\text{By the chain rule}} \cr
& - \frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( t \right)dt} } \right] = - \left[ {f\left( {g\left( x \right)} \right)g'\left( x \right)} \right] \cr
& - \frac{d}{{dx}}\left[ {\int_a^{g\left( x \right)} {f\left( t \right)dt} } \right] = - f\left( {g\left( x \right)} \right)g'\left( x \right) \cr} $$