Mathematics <Derivatives of Sin>

This is a discussion on Mathematics <Derivatives of Sin> within the A Brief History of Cprogramming.com forums, part of the Community Boards category; I have a problem that doesn't make sense to me. I will post the question, then the work I have ...

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    essence of digital xddxogm3's Avatar
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    Mathematics <Derivatives of Sin>

    I have a problem that doesn't make sense to me.
    I will post the question, then the work I have completed on it so far after.

    Question:
    Find the slope of the tangent line to the sine function at the origin. Compare this value with the number of complete cycles in the interval [0, 2p]. What can you conclude about the slope of the sine function sin(ax) at the origin?

    f(x)=sin(x)
    g(x)=sin(2x)

    my graphing calculator shows a deffinate steeper incline at the origin. This would make me assume that the Δy/Δx would be increasing as the number of cycles increase, but my calculator shows me different when calculating the the slope from the derivatives of these functions.

    f(x)=sin(x)
    f'(x)=cos(x)

    g(x)=sin(2x)
    g'(x)=cos(2x)

    when i evaluated f'(0) and g'(0) I get 1 for each of them. if the slope is visually identifiable as steeper, does that not mean the slope would also be analytically steeper? i know the domain for sin and cos are [-1,1], so i'm assuming this would reconfirm the calculators information, but the sloops do not look the same at point (0,0).

    Please if anyone can see where I'm confused or going wrong, point me to the light.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
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    C++ Developer XSquared's Avatar
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    g(x)=sin(2x)
    g'(x)=cos(2x)

    Actually, g'(x) = 2cos(2x).
    Naturally I didn't feel inspired enough to read all the links for you, since I already slaved away for long hours under a blistering sun pressing the search button after typing four whole words! - Quzah

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    essence of digital xddxogm3's Avatar
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    thanks for the input.
    i just noticed i was using the chain rule incorrectly on a few other problems also.
    that fixes it.
    ;0)

    anyway I have another question on the chain rule.

    knowing:
    dy/dx[f(u)] = f'(u)u'
    y=cotx
    y'=-csc^2(x)
    z=cscx
    z'=-cscxcotx
    sinx=(1/cscx)

    then shouldn't this be correct?
    f(x)=cotx/sinx
    f'(x)=-csc^2(x)(-cscxcotx)
    =csc^3(x)cotx
    =cotx/sin^3x

    the book says
    [-1-cos^2(x)]/sin^3(x)


    i'm assuming the book wants the answer in sin / cos.
    how would i get the [-1-cos^2(x)]
    i'm not aware of any trig convertion that would create this.
    any guidance would be helpful.
    Last edited by xviddivxoggmp3; 09-30-2005 at 11:24 PM.
    "Hence to fight and conquer in all your battles is not supreme excellence;
    supreme excellence consists in breaking the enemy's resistance without fighting."
    Art of War Sun Tzu

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    S Sang-drax's Avatar
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    Just convert it into sinuses and cosinuses and derive:

    f(x)=cotx/sinx = cos(x) / sin^2(x)

    f'(x) = [division rule] = (-sin^3(x) - 2sin(x)cos^2(x)) / sin^4(x) = (-1 - 2cos^2(x)) / sin^3(x)
    Last edited by Sang-drax : Tomorrow at 02:21 AM. Reason: Time travelling

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