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## Main Question or Discussion Point

Hey guys, I'm just doing some calculus practice over the winter break, and I came across this problem I've done a few times in a row, and I am either a) making the same mistake again, and again; or b) the book is wrong.

The problem is presented as such, with instructions to find the derivative:

g(x)=(2r sin rx + n)^p

With an answer from the back of the book:

g'(x) = [p(2r sin rx + n)^(p-1)] * (2r^2cos rx)

Here are the steps I use to solve:

g = u^p

u = (2r sin rx + n)

dg/du = pu^p-1 * u' : chain rule + product rule

u' = d/dr(2r * sin rx) + d/dr(n) : sum rule

d/dr(2r * sin rx) = (2r^2cos(rx)) + (2sinrx) : product rule

thus, u' = 2r^2cos(rx) + 2sinrx

thus,

dg/dr = [p(2r sin rx + n)^(p-1)] * (2r^2cosrx + 2sinrx) : just filling in the variables.

I'm "correct" right up until + 2 sinrx... that just shouldn't be there.

I've redone this several times, and like I said, I'm either misunderstanding something, or the book is wrong.

Thanks for any help.

The problem is presented as such, with instructions to find the derivative:

g(x)=(2r sin rx + n)^p

With an answer from the back of the book:

g'(x) = [p(2r sin rx + n)^(p-1)] * (2r^2cos rx)

Here are the steps I use to solve:

g = u^p

u = (2r sin rx + n)

dg/du = pu^p-1 * u' : chain rule + product rule

u' = d/dr(2r * sin rx) + d/dr(n) : sum rule

d/dr(2r * sin rx) = (2r^2cos(rx)) + (2sinrx) : product rule

thus, u' = 2r^2cos(rx) + 2sinrx

thus,

dg/dr = [p(2r sin rx + n)^(p-1)] * (2r^2cosrx + 2sinrx) : just filling in the variables.

I'm "correct" right up until + 2 sinrx... that just shouldn't be there.

I've redone this several times, and like I said, I'm either misunderstanding something, or the book is wrong.

Thanks for any help.