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Derivatives of Trig functions

  1. Jun 4, 2003 #1
    I didn't understand how they arrived at this step:


    can some one explain pls?
  2. jcsd
  3. Jun 5, 2003 #2


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    I doubt that anyone can explain how one step follows from the other because you haven't shown the whole steps.

    I assume what you have says that "lim as x -> 0 { 3 cos 3x (dy/dx) - 3 sin 3x (dy/dx)}" EQUALS something and that that follows from the fact that "lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }" equals something else. Unfortunately, you haven't told us what they equal.

    Since the only difference between the two that I can see is that the first above is multiplied by 3, I would suspect that

    "lim as x -> 0 { 3cos 3x (dy/dx) - 3sin3x (dy/dx) }= 3L"

    follows from
    "lim as x -> 0 { cos 3x (dy/dx) - sin3x (dy/dx) }= L" by multiplying the equation by 3. But of course, I can't be sure.
  4. Jun 5, 2003 #3
    they equal to -3 sin3x.

    I was just wondering how did they get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x...
  5. Jun 5, 2003 #4
    Please tell us exactly what it is that you (or "they") are trying to prove -- not just this one step, but the overall context.

    And, what is the entire expression or equation you are starting out with?
  6. Jun 5, 2003 #5


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    No, they DON'T "equal to -3 sin3x." They can't because each has a
    "lim as x->0". Once again, I can't say precisely what happened because you still haven't given us the whole thing but the obvious way to "get cos3x to be 3 cos 3x, sin 3x to be 3 sin 3x" is to multiply by 3!
  7. Sep 6, 2007 #6
    To get that by getting the derivative of 3x which is 3 then copy cos3x and you will get
    "3 cos 3x" same as "sin 3x to be 3 sin 3x". Not equal to -3 sin3x.
    Last edited: Sep 6, 2007
  8. Feb 20, 2008 #7
    if i understand what you mean, they just used the chain rule for derivatives.
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