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Derivative of sin(3x)

  1. Jun 5, 2005 #1
    Is the derivative of sin(3x) just 3sin(3x) because of the chain rule? IE, let u=3x, then 3sin(u) => 3sin(3x)


    Ok, I am pretty sure that is true. How about, [tex]

    \lim_{t\rightarrow 0} tln(t)


    What is the common approach for this problem?
    Last edited: Jun 5, 2005
  2. jcsd
  3. Jun 5, 2005 #2


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    It's not true because the derivative of sine is not itself

    For the limit, you might look for a minimum of the function. Does it have one? More than one? That should get you started.
  4. Jun 5, 2005 #3
    i agree with OlderDan.
    if you have that, then:
    so, sine is not the derivative of itself.
  5. Jun 5, 2005 #4


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    I think the derivative of sine is cosine,
    so the derivative of sin(3x) would be 3cos(3x)
  6. Jun 5, 2005 #5
    Bah, that is what I meant, sorry. Thanks, been a while since I did a derivative or a limit :cry:
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