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50th dy/dx

  1. Jul 7, 2004 #1
    The qeustion was what is the 50th dy/dx of F(x)=Cos(2x)
    this is what i did
    1st------dy/dx=-2Sin(2x)
    2nd------dy/dx=-4Cos(2x)
    3rd-------dy/dx=8Sin(2x)
    4th-------dy/dx=16Cos(2x)
    5th-------dy/dx=-32Sin(2x).....now that it reapeated itself its time to figure out the 50th. thats where im confused. My problem is with the negative signs. how can i catch that patteren..... :mad:
     
  2. jcsd
  3. Jul 7, 2004 #2

    AKG

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    Science Advisor
    Homework Helper

    If you think about it, it should be pretty clear. Differentiating cosine gives a negative, differentiating sine gives none. And you differentiate a cosine every other time.
    So, you start with
    -sin. Now, differentiating sin gives no negative, so nothing changes
    -cos. Cos changes things, so
    +sin
    +cos
    -sin
    -cos
    +
    +
    -
    -
    +
    +
    .
    .
    .

    Got it?
     
  4. Jul 7, 2004 #3
    hmm...yeah i see. Thanks!
     
  5. Jul 17, 2004 #4

    nta

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    change to cosine function

    I think you should use this way to solve your problem:
    1th dx/dy = -2sin2x=+2cos(2x+pi/2), so we obtain
    2th dx/dy = +4cos(2x+pi)
    3th dx/dy = +8cos(2x+3pi/2)
    Continueing, we obtain:
    i th dx/dy=+2^icos(2x+npi/2), from this, we can determine 50th dx/dy very fast and easy
     
  6. Jul 17, 2004 #5

    Galileo

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    You could try focusing on the even derivatives:

    0th: cos(2x)
    2nd: -4cos(2x)
    4th: 16cos(2x)
    etc.

    So you might guess for the (2n)th derivative:
    [tex](-1)^n4^ncos(2x)[/tex]

    and prove this with induction.
     
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