Answer
$$g'(x)=2r^2p\cos rx(2r\sin rx+n)^{p-1}$$
Work Step by Step
$$g'(x)=((2r\sin rx +n)^p)'=p(2r\sin rx+n)^{p-1}(2r\sin rx+n)'=
p(2r\sin rx+n)^{p-1}(2r\cos rx \cdot(rx)'+0)=
p(2r\sin rx+n)^{p-1}(2r\cos rx\cdot r)=2r^2p\cos rx(2r\sin rx+n)^{p-1}$$
