**Hint** : $(n)'=(2)'=(r)'=0.$ That simplifies things greatly.
Bear in mind also that you should have $$p(2ra^{rx}+n)^{p-1}\cdot\bigl((n)'+[(2)'\cdot(ra^{rx})+2\cdot[(r)'\cdot(a^{rx})+(r)\cdot a^{rx}\ln(a)\cdot r]]\bigr).$$ Note the added grouping symbols, and the fact that $(rx)'=r\
e rx.$
You will also make your life a great deal simpler by recalling/noting that when $f(x)$ is differentiable and $c$ is a constant, then $$\bigl(c\cdot f(x)\bigr)'=c\cdot f'(x).$$ This will save you multiple unnecessary uses of the product rule, since then we immediately see that $$(2ra^{rx}+n)'=(2ra^{rx})'+(n)'=2r(a^{rx})'+0,$$ and from there we continue in much the same way that you did.