Answer
$$\frac{9}{t}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dt}}\left( {\int_1^t {\frac{3}{x}dx} - \int_{{t^2}}^1 {\frac{3}{x}dx} } \right) \cr
& {\text{use the property of the derivatives}} \cr
& = \frac{d}{{dt}}\left( {\int_1^t {\frac{3}{x}dx} } \right) - \frac{d}{{dt}}\left( {\int_{{t^2}}^1 {\frac{3}{x}dx} } \right) \cr
& {\text{property }}\int_a^b {f\left( u \right)} du = - \int_b^a {f\left( u \right)} du \cr
& = \frac{d}{{dt}}\left( {\int_1^t {\frac{3}{x}dx} } \right) + \frac{d}{{dt}}\left( {\int_1^{{t^2}} {\frac{3}{x}dx} } \right) \cr
& {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr
& A\left( x \right) = \int_a^t {f\left( x \right)dx{\text{ }} \to {\text{ }}} A'\left( t \right) = \frac{d}{{dt}}\int_a^t {f\left( x \right)dx} = f\left( t \right) \cr
& {\text{use the chain rule for the second derivative}} \cr
& = \frac{3}{t} + \left( {\frac{3}{{{t^2}}}} \right)\frac{d}{{dt}}\left( {{t^2}} \right) \cr
& = \frac{3}{t} + \left( {\frac{3}{{{t^2}}}} \right)\left( {2t} \right) \cr
& {\text{simplify}} \cr
& = \frac{3}{t} + \frac{{6t}}{{{t^2}}} \cr
& = \frac{3}{t} + \frac{6}{t} \cr
& = \frac{9}{t} \cr} $$
