Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Is my book wrong?

  1. Dec 31, 2007 #1
    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

    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.
  2. jcsd
  3. Dec 31, 2007 #2
    g'(x) means dg(x)/dx. You seem to have calculated dg(r)/dr.
  4. Dec 31, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    It seems like the book is finding the derivative with respect to x, while you are finding it with respect to r.
  5. Dec 31, 2007 #4
    Oh wow. How right you are.
    I didn't even realize what the heck I was doing... I think I need a break :P Thank you!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook