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Math Help - Deriving Sinosoid Functions

  1. #1
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    Deriving Sinosoid Functions

    Could someone be so kind as to derive these functions for me? My answers are not what the back of the book has but i dont now where I go wrong so if you could show all your work I would appreciate it.

    y = cos^3x

    y = sin(x^3)

    f(x) = cos(5x-3)

    f(x) = sin^2xcos (\frac{x}{2})

    f(x) = cos^2(4x^2)

    g(x) = \frac{cosx}{cosx-sinx}

    Thanks
    Chris
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by ffezz View Post
    Could someone be so kind as to derive these functions for me? My answers are not what the back of the book has but i dont now where I go wrong so if you could show all your work I would appreciate it.

    y = cos^3x

    y = sin(x^3)

    f(x) = cos(5x-3)

    f(x) = sin^2xcos (\frac{x}{2})

    f(x) = cos^2(4x^2)

    g(x) = \frac{cosx}{cosx-sinx}

    Thanks
    Chris
    What do you mean derive? Differentiate these sinusoidal functions?
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  3. #3
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    I think it's just about contemplating the derivative of each one.
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  4. #4
    Moo
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    Hi,

    Quote Originally Posted by ffezz View Post
    Could someone be so kind as to derive these functions for me? My answers are not what the back of the book has but i dont now where I go wrong so if you could show all your work I would appreciate it.
    It's better for you to show your working so that we can more easily go through your mistakes and explain what's wrong
    I guess we learn more from mistakes than full solutions..
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  5. #5
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    ok heres my work

    yes i did mean differentiate =)


    f(x) = cos^2(4x^2)

    f'(x) = 2cos(4x^2)-sin(4x^2)(8x)

    = -16xcos(4x^2)-sin(4x^2)
    Thats as far as I can get and this appears to be a f(g(h(x))) function. The answer is cut off for this one so I dont know if its right or no

    now this one I have no idea

     sin^2xcos(\frac{x}{2})

    f'(x) = 2sinxcosxcos(\frac{x}{2})sin^2x-sin(\frac{x}{2})

    I dont know where to go from there

    I just moved on to the next question and it is even more difficult.

    "Dtetermine the equation of the tangent to

    y = 2 + cos2x at x = \frac{5pi}{6}

    the answer is appearntly \sqrt{3x} + 3 - \frac{5\sqrt{3}}{6}pi
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  6. #6
    Moo
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    Ok, that's far better !

    One thing, essential thing you have to keep in mind for this exercise is the chain rule :

    [f(g(x))]'=g'(x) \cdot f'(g(x))

    From this rule, you can do funny things, especially this one :

    [f(g(h(x)))]'=[f(k(x))]', with k(x)=g(h(x))

    Using the chain rule :

    [f(k(x))]'=k'(x) \cdot f'(k(x))

    Now, what is k'(x) ? k'(x)=[g(h(x))]'=h'(x) \cdot g'(h(x))

    So \boxed{[f(g(h(x)))]'=h'(x) \cdot g'(h(x)) \cdot f'(g(h(x)))}



    Quote Originally Posted by ffezz View Post
    yes i did mean differentiate =)


    f(x) = cos^2(4x^2)

    f'(x) = 2cos(4x^2)-sin(4x^2)(8x)

    = -16xcos(4x^2)-sin(4x^2)
    Thats as far as I can get and this appears to be a f(g(h(x))) function. The answer is cut off for this one so I dont know if its right or no

    now this one I have no idea

     sin^2xcos(\frac{x}{2})

    f'(x) = 2sinxcosxcos(\frac{x}{2})sin^2x-sin(\frac{x}{2})

    I dont know where to go from there

    I just moved on to the next question and it is even more difficult.

    "Dtetermine the equation of the tangent to

    y = 2 + cos2x at x = \frac{5pi}{6}

    the answer is appearntly \sqrt{3x} + 3 - \frac{5\sqrt{3}}{6}pi
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