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Differentiation problem

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    If g(x) + x sin g(x) = x^2 find g'(0)

    2. Relevant equations



    3. The attempt at a solution
    At this point I have tried a few things but hit deadends. Any help would be appreciated.
     
  2. jcsd
  3. Oct 13, 2009 #2

    lanedance

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    Homework Helper

    try using implict differentiation, then sub in for x = 0
     
  4. Oct 13, 2009 #3

    Mark44

    Staff: Mentor

    Or just differentiate.

    Show us what you have tried...
     
  5. Oct 13, 2009 #4
    my first step took me to this:
    g'(x) + xcos g'(x) + sin g(x) = 2x
    then i set it equal to g'(x)
    g'(x)= 2x- sin g(x)/xcos(1)
    I would then plug in 0 but that makes the denominator 0
     
  6. Oct 13, 2009 #5

    Mark44

    Staff: Mentor

    The first, third, and fourth terms are correct, but the second is wrong. To differentiate x sin(g(x)) you need the product rule and the chain rule.
    d/dx(x sin(g(x))) = x*d/dx(sin(g(x)) + 1* sin(g(x))

    d/dx(sin(g(x)) is NOT cos(g'(x)). That's not how the chain rule works.

    What should you get with d/dx(f(g(x))?
     
  7. Oct 13, 2009 #6
    d/dx f(g(x)) would be f'(x)g(x)*g'(x)
     
  8. Oct 13, 2009 #7

    Mark44

    Staff: Mentor

    Yes. Now what is d/dx(sin(g(x))?
     
  9. Oct 13, 2009 #8
    cos(g(x))*g'(x)
     
  10. Oct 13, 2009 #9

    Mark44

    Staff: Mentor

    Yes. Now put this back your differentiation problem for the part that was incorrect.
     
  11. Oct 13, 2009 #10
    alright i ended up getting
    g'(x)= 2x- sin g(x)/(1+cos g(x)
    so then i plug in the 0 but what is g(0)
     
  12. Oct 13, 2009 #11

    lanedance

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    Homework Helper

    have a look at your original equation, when x = 0
     
  13. Oct 13, 2009 #12
    not quite my friends...there is a solution to this one. dig more.
     
  14. Oct 13, 2009 #13

    lanedance

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    yeah i think we already outlined it
     
  15. Oct 14, 2009 #14
    First of all theres something missing in this equation: the x in front of cos g(x).
     
  16. Oct 14, 2009 #15
    You should obtain an equation in the form of: g'(x)=[2x-sin g(x)]/[1+xcos g(x)].
     
  17. Oct 14, 2009 #16
    Now, understanding what the problem really asks for, it is asking for g'(0), meaning what is the value of g'(x) when x=0.
     
  18. Oct 14, 2009 #17
    You can substitute 0 now for x, you will end up getting: g'(0) = -sin g(x)<----this solves the problem.
     
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