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Derivative using chain rule

  1. Apr 8, 2008 #1
    I am trying to find the first and second derivative using the chain rule of the following:

    u sin(x^2)

    This is what I have but I don't think it is correct. Can someone pls let me know?

    first derivative: u * 2x cos(x^2) + sin(x^2) u'

    second derivative:
    u * 2( x * -2sin(x^2) + cos(x^2)) + 2xcos(x^2)* u' + sin(x^2)*u" + u'* 2xcos(x^2)

    Any help thanks
     
  2. jcsd
  3. Apr 8, 2008 #2

    jamesrc

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    Gold Member

    That's pretty close: the first term (for your second derivative) is not correct but I think if you take a second look you will figure it out (you just missed one detail). Otherwise, you can combine like terms to clean it up a bit, but it's still a long expression.
     
  4. Apr 8, 2008 #3
    I don't get it :(
     
  5. Apr 8, 2008 #4
    i am assuming u is an other function right?

    Think of it this way

    [tex] [(u2x)cos(x^2)]'[/tex] in other words think first of (u2x) as a single function, so you can differentiate this using the product rule for (u2x) and
    cos(x^2) right? after that apply again the product rule wherever you end up with (u2x)' and you'll be fine.
     
  6. Apr 8, 2008 #5
    This is what I get

    2ucos(x^2)-4x^2sin(x^2)+4xcos(x^2)u'+sin(x^2)u"

    Is this correct?
    Thanks
     
  7. Apr 8, 2008 #6

    You have this question on another post as well, and it looked fine to me as well there.
    Well i take this back, you are short of an x in there.
    you should have -2xsin(x^2) somewhere at the beggining, on your original result, at fpost #1.
     
    Last edited: Apr 8, 2008
  8. Apr 8, 2008 #7
    This looks fine.
     
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