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Very easy question

  1. Sep 27, 2006 #1

    DB

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    simple question, if the derivative of cos(x) as x->0 is 1, the is the derivative of cos(3x) as x->0 = 1 aswell?
     
  2. jcsd
  3. Sep 27, 2006 #2

    AKG

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    The derivative of cos(x) as x->0 is -sin(x) as x->0 is -sin(0) is 0, not 1. But simply cos(x) as x->0 is cos(0) = 1. the derivative of cos(3x) as x->0 = -3sin(3x) as x->0 = -3sin(3*0) = -3sin(0) = -3*0 = 0. cos(3x) as x->0 = cos(3*0) = cos(0) = 1. The derivative of cos(x) as x->a = -sin(x) as x->a = -sin(a). The derivative of cos(3x) as x->a = -3sin(3x) as x->a = -3sin(3a).
     
    Last edited: Sep 28, 2006
  4. Sep 28, 2006 #3

    mathman

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    The derivative of cos(x) is -sin(x).
     
  5. Sep 28, 2006 #4

    AKG

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    Yeah, I knew that. :redface: Edited.
     
  6. Oct 1, 2006 #5
    Think about what a derivative means. This should be plenty enough to give you the answer you're looking for.
     
  7. Oct 4, 2006 #6

    HallsofIvy

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    The derivative of sin x goes to 1 as x goes to 0. No, the derivative of sin(3x) does not go to 1. It goes to 3. Do you see why?
     
  8. Oct 4, 2006 #7
    Curious... it looks like you're looking for a limit. I wasn't aware that derivitaves "approached" anything.
     
  9. Oct 4, 2006 #8

    HallsofIvy

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    The derivative of a function of x is a function of x and can approach a limit as well as any function.
     
  10. Oct 4, 2006 #9
    The derivative tells you how the result will change with a change in the x. If it were to linearly change, the derivative would say how y changes when 1 is added to x. So, if you have cos(3x), the derivative will have to be 3 times greater than cosx, because if you raise x by an amount (say, 1), you are raising the quantity inside the cos by 3 times that amount. It's the same concept as the derivative (slope) of y=x and y=3x.

    This is how the chain rule can be derived....
     
  11. Oct 5, 2006 #10
    just U-substitute u=3x if you must and apply chain rule. dy/dx=dy/du*du/dx
     
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