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Math Help - Trig differentation

  1. #1
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    Trig differentation

    How do you derive this equation:

    y = sin ^ 3 (cosx^3)
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  2. #2
    MHF Contributor matheagle's Avatar
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    This is the chain rule if the cosine is inside the sine, which what this seems to be.
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  3. #3
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    Quote Originally Posted by mathamatics112 View Post
    How do you derive this equation:

    y = sin ^ 3 (cosx^3)
    You need to use the chain rule several times:

    y=\sin^3\cos x^3=\left(\sin\cos x^3\right)^3

    \Rightarrow\frac{dy}{dx}=3\sin^2\cos x^3\cdot\frac d{dx}\left[\sin\cos x^3\right]

    =3\sin^2\cos x^3\left(\cos\cos x^3\cdot\frac d{dx}\left[\cos x^3\right]\right)

    And so on. Can you continue?
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  4. #4
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    y= sin^3( cos x^3)
    = 3 sin ^ 2 ( cos x^3)(coz(cosx^3))(-sinx^3)(3x^2)
    = 9 x^2(sin^2)(cos(cos x^3))(-sin x^3))


    are my steps right?
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  5. #5
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    Quote Originally Posted by mathamatics112 View Post
    y= sin^3( cos x^3)
    = 3 sin ^ 2 ( cos x^3)(coz(cosx^3))(-sinx^3)(3x^2)
    = 9 x^2(sin^2)(cos(cos x^3))(-sin x^3))
    Aren't you missing something in that last line?
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  6. #6
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    i'm not seeing it
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  7. #7
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by mathamatics112 View Post
    i'm not seeing it
    Did you look at the part that I highlighted? You have \text{``}\sin^2\text{''} sitting by itself. You are taking the sine-squared of what?
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