Artificial intelligent assistant

chain rule: $g(x)=(2ra^{rx}+n)^p$ looking for my mistake and I can't find it. here is the layout: $g(x)=(2ra^{rx}+n)^p$ $p(2ra^{rx}+n)^{p-1} (2ra^{rx}+n)'$ $p(2ra^{rx}+n)^{p-1} (2ra^{rx})'+(n)'$ $p(2ra^{rx}+n)^{p-1} (n)'+[(2)' (ra^{rx}) + (ra^{rx})' (2)]$ $p(2ra^{rx}+n)^{p-1} (n)'+[(2)' (ra^{rx}) + (2) [(r)'(a^{rx})+(a^{rx})'(r)]]$ $p(2ra^{rx}+n)^{p-1} (n)'+[(2)' (ra^{rx}) + (2) [(r)'(a^{rx})+(r)a^{rx}\ln(a)(rx)']]$ which when finally unpacked should be: $p(2ra^{rx}+n)^{p-1} (n)'+[(2)'(ra^{rx}) + (2) [(r)' (a^{rx})+(r)a^{rx}\ln(a)rx]]$ the answer i got was: $$p(2ra^{rx}+n)^{p-1}+2a^{rx}+2r^{2}xa^{rx}\ln(a)$$ the answer says: $$2r^{2}p(\ln a)(2ra^{rx}+n)^{p-1}a^{rx}$$

**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.

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