Answer
$g'(α)=(5^{-\frac{α}{2}})((-\frac{1}{2}\log 5)(sin(2α))+2cos(2α))$
Work Step by Step
$g(α)= 5^{-\frac{α}{2}}sin2α$
$u=5^{-\frac{α}{2}}$
$u’=(-\frac{1}{2}\ln 5)5^{-\frac{α}{2}}$
$v=sin2α$ $v’=2cos2α$
$g’(α)=u’v+uv’$
$=((-\frac{1}{2}\ln 5)5^{-\frac{α}{2}})(sin(2α))+(5^{-\frac{α}{2}})(2cos(2α))$
$=(5^{-\frac{α}{2}})((-\frac{1}{2}\log 5)(sin(2α))+2cos(2α))$