Answer
$A'\left( w \right) = - {e^{1 + {w^2}}}$
Work Step by Step
$$\eqalign{
& A\left( w \right) = \int_w^{ - 1} {{e^{1 + {t^2}}}} dt \cr
& {\text{Use the integral property }}\int_a^b {f\left( x \right)} dx = - \int_b^a {f\left( x \right)} dx{\text{ }} \cr
& A\left( w \right) = - \int_{ - 1}^w {{e^{1 + {t^2}}}} dt \cr
& {\text{Differentiate both sides with respect to }}w \cr
& A'\left( w \right) = \frac{d}{{dw}}\left[ { - \int_{ - 1}^w {{e^{1 + {t^2}}}} dt} \right] \cr
& A'\left( w \right) = - \frac{d}{{dw}}\left[ {\int_{ - 1}^w {{e^{1 + {t^2}}}} dt} \right] \cr
& {\text{Use Part 1 of the Fundamental Theorem of calculus}} \cr
& g\left( w \right) = \int_a^w {f\left( t \right)} dt,{\text{ }}a \leqslant w \leqslant b,{\text{ }} \Rightarrow {\text{ }}g'\left( w \right) = f\left( w \right) \cr
& {\text{Let }}f\left( t \right) = {e^{1 + {t^2}}} \cr
& f\left( w \right) = {e^{1 + {w^2}}} \cr
& {\text{Therefore}}{\text{,}} \cr
& A'\left( w \right) = - \frac{d}{{dw}}\left[ {\int_{ - 1}^w {{e^{1 + {t^2}}}} dt} \right] \cr
& A'\left( w \right) = - {e^{1 + {w^2}}} \cr} $$
