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Math Help - [SOLVED] Derivatives of cos and sin

  1. #1
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    Post [SOLVED] Derivatives of cos and sin

    I need help to get the derivative of the function:

    2cosx.cos(9x)

    and then the derivative of the derivative. Explanation of how much appreciated.
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  2. #2
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    Quote Originally Posted by Buzzins View Post
    I need help to get the derivative of the function:

    2cosx.cos(9x)

    and then the derivative of the derivative. Explanation of how much appreciated.
    2cosx.cos(9x)

    In trigonometry there is a formula to convert product to sum of trig. functions

    2cos(A)cos(B) = cos(A+B) + cos(A-B)

    Using this formula you can write this as

    cos(10x) + cos(8x)

    Now find the derivative.
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  3. #3
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    thats my problem. I have no idea where to start or what to do. Trig's not my thing
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  4. #4
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    Quote Originally Posted by Buzzins View Post
    thats my problem. I have no idea where to start or what to do. Trig's not my thing
    Do you know the derivative of cos(mθ) ?
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  5. #5
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    as in the derivative of cosx is -sinx ?
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  6. #6
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    Equations and relations we're going to use here:
    (1) Product rule: [f(x)*g(x)]' = f(x)*g'(x) + f'(x)*g(x)
    (2) Derivative of sin/cos: sin'(x) = cos(x) , cos'(x) = -sin(x)
    (3) Chain rule: [g(f(x))]' = g'(f(x))*f'(x). ie: sin'(9x)=cos(9x)*9

    First derivative:

    It will be convenient to define:
    f(x) = 2cos(x)
    g(x) = cos(9x)

    Derivative of a product of two functions:
    (eq.1) (f(x)*g(x))' = f(x)*g'(x) + f'(x)*g(x)
    In this case:
    f'(x) = (2cos(x))' = 2*cos'(x) = 2*(-sin(x)) = -2sin(x)
    g'(x) = (cos(9x))' = -sin(9x)*9 = -9sin(9x)

    We plug those back in eq.1:
    (2cos(x)cos(9x))' = 2cos(x)*(-9sin(9x))+(-2sin(x))*cos(9x)) = ..
    .. = -18cos(x)sin(9x) - 2sin(x)cos(9x)

    Second derivative:

    The derivative of -18cos(x)sin(9x):
    define: f(x) = -18cos(x) , g(x) = sin(9x)
    now
    f'(x) = -18cos'(x) = -18*(-sin(x)) = 18sin(x)
    g'(x) = sin'(9x) = 9cos(9x)

    and so, like we did above:
    (-18cos(x)sin(9x))' = -18cos(x)*9cos(9x) + 18sin(x)*sin(9x) = ..
    .. = -162cos(x)cos(9x) + 18sin(x)sin(9x)


    The derivative of -2sin(x)cos(9x):
    define: f(x) = -2sin(x) , g(x) = cos(9x)
    and now:
    f'(x) = -2sin'(x) = -2cos(x)
    g'(x) = cos'(9x) = -9sin(9x)
    -2sin(x)cos(9x) = (-2sin(x))(-9sin(9x))+(-2cos(x)cos(9x)) = ..
    .. = 18sin(x)sin(9x) - 2cos(x)cos(9x)


    So finally, the derivative of -18cos(x)sin(9x)-2sin(x)cos(9x) is:

    -162cos(x)cos(9x) + 18sin(x)sin(9x) + 18sin(x)sin(9x) - 2cos(x)cos(9x) = ..
    .. = 36sin(x)sin(9x) - 164cos(x)cos(9x)
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  7. #7
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    thanks mate. The help was much appreciated
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