Trig/Quotient Rule

Discussion in 'Undergraduate Math' started by James Stein, Oct 1, 2004.

  1. James Stein

    James Stein Guest

    Two point post, actually. The first is a question; does anyone know of a
    good site covering basic trig, up to trig identities and trig functions?
    I've always been bad with that, and I was hoping to get some review in (my
    textbook doesn't cover it well.)

    The second is the following:

    f(x) = 5sin(-3x) / 1 + cosx.
    Find f ' (3).

    So, using quotient rule:
    (1 + cos x) (5cos(-3x)) - (5sin(-3x))(-sinx)
    -------------------------------------------
    (1 + cosx)^2

    I'm really not sure how to foil it, because I'm not sure what happens with
    5cos(-3x) * cos x. Does it become 5cos^2(-3x), or 5cos(-3x^2) or
    5cos(-3x)cosx?

    Just plugging the three into the formula as it stands doesn't get me the
    right answer, from what I see.

    Also, I couldn't think of any trig ID's to simplify the issue.
     
    James Stein, Oct 1, 2004
    #1
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  2. James Stein

    David Moran Guest

    You're close, but check your differentiation of 5sin(-3x). I don't think
    there's any way of simplifying this.

    Dave
     
    David Moran, Oct 1, 2004
    #2
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  3. Quotient rule is:

    d(uv)/dx = (v du/dx - u dv/dx)/v^2

    u(x) = 5sin(-3x) = -5sin(3x) [I don't like having negatives in
    the argument of trig functions]

    v(x) = 1 + cos(x)

    du/dx = (-5)(3)cos(3x) = -15cos(3x) [you forgot about the 3]

    dv/dx = -sin(x)

    so

    df/dx = -15[1 + cos(x)]cos(3x) - 5sin(3x)sin(x)
    ---------------------------------------
    [1 + cos(x)]^2
    Why do you want to FOIL it? Just plug in x=3.
    No, it doesn't. Why do you think it might (I'm not being
    sarcastic -- I'm trying to see why you think that's even
    a possibility)?
    See above.
    That's it. Aside from using the
    cos(a)cos(b) = [cos(a+b) + cos(a-b)]/2
    formula to write it as a sum of individual
    cosines, that's about all you can do with it.
     
    Rich Carreiro, Oct 1, 2004
    #3
  4. James Stein

    James Stein Guest

    Forget about it? I don't even know where it came from. Is that a chain rule
    multiplying by d/dx 3x?

    I have a very shaky grasp of basic trig. I manage to pull off high 90's on
    my calc tests anyway, but mostly through rote memorization of
    problem-solving strategies and certain recurrent 'tricks' - not through
    understanding of the fundamentals. Which is unfortunate, and something I'm
    working to correct.
     
    James Stein, Oct 1, 2004
    #4
  5. Yes, it's because of the chain rule.

    Remember, d[f(u(x))]/dx = (df/du)(du/dx)

    For sin(3x), f(u) = sin(u), u(x) = 3x, so
    d(sin(3x))/dx = d[sin(u)]/du * d(3x)/dx
    = cos(u) * 3
    = 3cos(u)
    = 3cos(3x)
     
    Rich Carreiro, Oct 1, 2004
    #5
  6. James Stein

    Stan Brown Guest

    (ahem) I venture to offer my Trig Without Tears, at
    http://oakroadsystems.com/twt/

    --
    Stan Brown, Oak Road Systems, Tompkins County, New York, USA
    http://OakRoadSystems.com
    A: Maybe because some people are too annoyed by top-posting.
    Q: Why do I not get an answer to my question(s)?
    A: Because it messes up the order in which people normally read
    text.
    Q: Why is top-posting such a bad thing?
     
    Stan Brown, Oct 1, 2004
    #6
  7. James Stein

    Virgil Guest

    One point question on notation. May one presume that your function is
    equivalent to
    f(x) = 5 sin(-3 x) / (1 + cos(x))= 5*sin(-3*x) / (1 + cos(x))
    ?

    It is customary in ASCII representations to either leave a space or use
    something like "*" to indicate multiplications, to enclose the arguments
    of all functions, including trig functions, in parentheses, and to
    enclose divisors or denominators of more than one term or factor in
    parentheses, as I have done.
     
    Virgil, Oct 1, 2004
    #7
  8. = 5sin(-3x) + cos x
    f'(x) = -15cos(-3x) - sin x
     
    William Elliot, Oct 1, 2004
    #8
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