Answer
$$ - \frac{{2x}}{{{x^4} + 1}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\int_{{x^2}}^{10} {\frac{{dz}}{{{z^2} + 1}}} \cr
& {\text{Property }}\int_a^b {f\left( u \right)} du = - \int_b^a {f\left( u \right)} du \cr
& = - \frac{d}{{dx}}\int_{10}^{{x^2}} {\frac{{dz}}{{{z^2} + 1}}} \cr
& {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr
& A\left( x \right) = \int_a^x {f\left( t \right)dt{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right) \cr
& {\text{the upper limit is }}{x^2},{\text{ so is needed to apply the chain rule}} \cr
& - \frac{d}{{dx}}\int_{10}^{{x^2}} {\frac{{dz}}{{{z^2} + 1}}} = - \frac{{dy}}{{dx}} = - \frac{{dy}}{{du}}\frac{{du}}{{dx}} \cr
& {\text{with }}u = {x^2}{\text{ and }}a = 10 \cr
& = - \left( {\frac{d}{{du}}\int_{10}^u {\frac{{dz}}{{{z^2} + 1}}} } \right)\frac{d}{{dx}}\left( {{x^2}} \right) \cr
& = - \left( {\frac{d}{{du}}\int_{10}^u {\frac{{dz}}{{{z^2} + 1}}} } \right)\left( {2x} \right) \cr
& {\text{by the fundamental theorem }}\left( {{\text{part 1}}} \right) \cr
& = - \left( {\frac{1}{{{u^2} + 1}}} \right)\left( {2x} \right) \cr
& = - \left( {\frac{1}{{{{\left( {{x^2}} \right)}^2} + 1}}} \right)\left( {2x} \right) \cr
& = - \frac{{2x}}{{{x^4} + 1}} \cr} $$
