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Math Help - derivative.. problem

  1. #1
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    derivative.. problem

    Hello
    can someone please show me how to do this sum correct
    I need to find the derivative of x^-4(1-X^-4)^3

    the correct answer is in yellow at the bottom,my answer is above it

    please see attached photo thanks
    Attached Thumbnails Attached Thumbnails derivative.. problem-satmat.jpg  
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  2. #2
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    Quote Originally Posted by wolfhound View Post
    Hello
    can someone please show me how to do this sum correct
    I need to find the derivative of x^-4(1-X^-4)^3
    x^{-4}(1-x^{-4})^3 = \frac{(x^4-1)^3}{x^{16}}

    \frac{d}{dx}\left[\frac{(x^4-1)^3}{x^{16}}\right]

    \frac{x^{16} \cdot 3(x^4-1)^2 \cdot 4x^3 - (x^4-1)^3 \cdot 16x^{15}}{x^{32}}

    \frac{4x^{15}(x^4-1)^2 [3x^4 - 4(x^4-1)]}{x^{32}}

    \frac{4x^{15}(x^4-1)^2 [4-x^4]}{x^{32}}

    \frac{4(x^4-1)^2 (4-x^4)}{x^{17}}
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  3. #3
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    Quote Originally Posted by wolfhound View Post
    Hello
    can someone please show me how to do this sum correct
    I need to find the derivative of x^-4(1-X^-4)^3

    the correct answer is in yellow at the bottom,my answer is above it

    please see attached photo thanks
    Looks like an application of the product and chain rule

    x^{-4}(1-x^{-4})^3

    u = x^{-4} \: \: \rightarrow \: \: u' = -4x^{-3}

    v = (1-x^{-4})^3 \: \: \rightarrow \:\: v' = 3(1-x^{-4})^2 \cdot 4x^{-3} = 12x^{-3}(1-x^{-4})^2

    y' = -4x^{-3}(1-x^{-4})^3 + 12x^{-7} \cdot (1-x^{-4})^2

    y' = -4x^{-3}[(1-x^{-4})^3 + 3x^{-4}(1-x^{-4})^2]


    Why they want you to use the binomial expansion is beyond me but it's as follows up to x^3


    (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3
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  4. #4
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    Quote Originally Posted by e^(i*pi) View Post
    Looks like an application of the product and chain rule

    x^{-4}(1-x^{-4})^3

    u = x^{-4} \: \: \rightarrow \: \: u' = -4x^{-3}

    v = (1-x^{-4})^3 \: \: \rightarrow \:\: v' = 3(1-x^{-4})^2 \cdot 4x^{-3} = 12x^{-3}(1-x^{-4})^2

    y' = -4x^{-3}(1-x^{-4})^3 + 12x^{-7} \cdot (1-x^{-4})^2

    y' = -4x^{-3}[(1-x^{-4})^3 + 3x^{-4}(1-x^{-4})^2]


    Why they want you to use the binomial expansion is beyond me but it's as follows up to x^3


    (1+x)^n = 1+nx+\frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3
    Hello, should it not be  u' = -4x^{-5}
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  5. #5
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    Quote Originally Posted by wolfhound View Post
    Hello, should it not be  u' = -4x^{-5}
    yes it should.
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