Answer
$$ - {e^{{x^2}}}$$
Work Step by Step
$$\eqalign{
& \frac{d}{{dx}}\int_x^1 {{e^{{t^2}}}dt} \cr
& {\text{property }}\int_a^b {f\left( u \right)} du = - \int_b^a {f\left( u \right)} du \cr
& = - \frac{d}{{dx}}\int_1^x {{e^{{t^2}}}dt} \cr
& {\text{using The Fundamental Theorem }}\left( {{\text{part 1}}} \right) \cr
& A\left( x \right) = \int_a^x {f\left( t \right)dt{\text{ }} \to {\text{ }}} A'\left( x \right) = \frac{d}{{dx}}\int_a^x {f\left( t \right)dt} = f\left( x \right) \cr
& then \cr
& - \frac{d}{{dx}}\int_1^x {{e^{{t^2}}}dt} = - {e^{{{\left( x \right)}^2}}} \cr
& {\text{Simplify}} \cr
& = - {e^{{x^2}}} \cr} $$
