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Two Derivatives.

  1. Jun 6, 2003 #1
    Here's what i got.
    f'(x) = pi * d/dx (x) + d/dx (cos^2 pi*x)-1
    = pi + [ (- (d/dx cos^2 (pi*x))/(cos^4(pi*x)) ]
    = pi + [ (- (d/dx cos(pi*x) * (2 * cos(pi * x))/(cos^4(pi*x)) ]
    = pi + [ (pi * sin x * 2cos(pi * x)) / ... ]
    = pi + [ (2 * pi * sin x) / cos^3 (pi * x) ]

    Thus f'(1/4) = 7.5251
     
  2. jcsd
  3. Jun 6, 2003 #2
    #2 question:


    f'(x) = d/dx ((1/x^2) - 2)/(1/x^2) * (2 * ((1/x^2) - 2)/(1/x^2))
    = [2((1/x^2) - 2)/(1/x^2)] * [((1/x^2)(-2/x^3) - ((1/x^2)-2)(-2/x^3))/((1/x^2)^2)]
    = ... * [ -4/x^7 ]
    = -8(1 - 2x^2)/x^7

    Thus, f'(-1) = -8.

    Please correct my mistakes. Thanks.
     
  4. Jun 6, 2003 #3

    Tom Mattson

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    I don't have a calculator handy, but your derivative is correct.

    For the derivative, I get

    f'(x)=-8x+16x3

    I rewrote the function as follows:

    f(x)=(1/x2-2)2x4

    To get the last factor, I noted that 1/(1/x2) is simply x2. Squaring that gives x4.

    edit: It would be best to expand the function before differentiating, as follows:

    f(x)=(1/x4-4/x2+4)x4

    Now distribute the x4

    f(x)=(1-4x2+4x4)

    Doesn't look as nasty now, does it? :smile:
     
    Last edited: Jun 6, 2003
  5. Jun 6, 2003 #4
    You sure you got your #2 derivative right?

    It took me two times to re-derive it and everytime I get the same answer.
     
  6. Jun 6, 2003 #5
    Should be this on the fourth line:
    = π + π sin(π x) 2cos(π x)/cos4(π x)
    i.e. Missed a π inside the sin(x).

    And the final answer:
    f'(x) = π + 2π sec2(π x)tan(πx)
    f'(1/4) ≈ 15.708

    For your second question, you surprisingly did get the correct value for f'(-1), but the function you found is in general wrong:
    f(x) = [(1/x2-2) / (1/x2)]2
    f(x) = [(1/x2-2) *x2]2
    f(x) = (1/x2-2)2*x22
    f(x) = (1/x4-4/x2+4)x4
    f(x) = 1-4x2+4x4
    (as Tom derived)

    From this it far easier to find f'(x):
    f'(x)= -8x+16x3
    f'(x) = 8x(2x2-1)
    f'(-1) = -8
    This is similar to be not the same as the function you derived.
     
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