Answer
$A'\left( r \right) = \left( {\frac{{4{r^2} + 1}}{{2\sqrt r }}} \right){e^{{r^2} + 1}}$
Work Step by Step
$$\eqalign{
& A\left( r \right) = \sqrt r \cdot {e^{{r^2} + 1}} \cr
& {\text{Differentiate}} \cr
& A'\left( r \right) = \frac{d}{{dr}}\left[ {\sqrt r \cdot {e^{{r^2} + 1}}} \right] \cr
& {\text{Use the product rule for derivatives}} \cr
& A'\left( r \right) = \sqrt r \frac{d}{{dr}}\left[ {{e^{{r^2} + 1}}} \right] + {e^{{r^2} + 1}}\frac{d}{{dr}}\left[ {\sqrt r } \right] \cr
& {\text{Use the chain rule for }}\frac{d}{{dr}}\left[ {{e^{{r^2} + 1}}} \right] \cr
& A'\left( r \right) = \sqrt r \left( {{e^{{r^2} + 1}}} \right)\frac{d}{{dr}}\left[ {{r^2} + 1} \right] + {e^{{r^2} + 1}}\frac{d}{{dr}}\left[ {\sqrt r } \right] \cr
& {\text{Compute derivatives}} \cr
& A'\left( r \right) = \sqrt r \left( {{e^{{r^2} + 1}}} \right)\left( {2r} \right) + {e^{{r^2} + 1}}\left( {\frac{1}{{2\sqrt r }}} \right) \cr
& {\text{Multiplying}} \cr
& A'\left( r \right) = 2r\sqrt r {e^{{r^2} + 1}} + \frac{{{e^{{r^2} + 1}}}}{{2\sqrt r }} \cr
& {\text{Factor and simplify}} \cr
& A'\left( r \right) = \left( {2r\sqrt r + \frac{1}{{2\sqrt r }}} \right){e^{{r^2} + 1}} \cr
& A'\left( r \right) = \left( {\frac{{4{r^2} + 1}}{{2\sqrt r }}} \right){e^{{r^2} + 1}} \cr} $$
