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Math Help - Strange derivative question

  1. #1
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    Strange derivative question

    We are given that:

    f(x)=\displaystyle{\frac{(x-4)^2}{2x^\frac{1}{2}}}

    Show that:

    f'(x)=\displaystyle{\frac{3x^2-8x-16}{4x^\frac{3}{2}}}


    I've tried to figure this out and it just won't work!

    Can anybody help please?
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  2. #2
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    Hello, anthmoo!

    It's not "strange" . . . You're making an error or two.


    Given: . f(x)\:=\:\frac{(x-4)^2}{2x^{\frac{1}{2}}}

    Show that: . f'(x)\:=\:\frac{3x^2-8x-16}{4x^{\frac{3}{2}}}

    Quotient Rule: . f'(x)\:=\:\frac{2x^{\frac{1}{2}}\cdot2(x-4) - (x-4)^2\cdot x^{-\frac{1}{2}}}{4x} \;=\;\frac{4x^{\frac{1}{2}}(x-4) - x^{-\frac{1}{2}}(x-4)^2}{4x}


    Multiply top and bottom by x^{\frac{1}{2}}

    . . \frac{x^{\frac{1}{2}}}{x^{\frac{1}{2}}}\cdot\frac{  4x^{\frac{1}{2}}(x-4) - x^{-\frac{1}{2}}(x-4)^2}{4x} = \;\frac{4x(x-4) - (x-4)^2}{4x^{\frac{3}{2}}} \;=\;\frac{4x^2 - 16x - (x^2 - 8x + 16)}{4x^{\frac{3}{2}}}


    Therefore: . f'(x)\;=\;\frac{3x^2 - 8x - 16}{4x^{\frac{3}{2}}}

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  3. #3
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    Quote Originally Posted by anthmoo View Post
    We are given that:

    f(x)=\displaystyle{\frac{(x-4)^2}{2x^\frac{1}{2}}}

    Show that:

    f'(x)=\displaystyle{\frac{3x^2-8x-16}{4x^\frac{3}{2}}}


    I've tried to figure this out and it just won't work!
    Can anybody help please?
    Hello Anthmoo,

    use quotient rule:

    f'(x)=\displaystyle{\frac{2x^\frac{1}{2}\cdot 2(x-4)-(x-4)^2\cdot 2 \cdot \frac{1}{2}x^{-\frac{1}{2}}}{4x}} = \displaystyle{\frac{2x^\frac{1}{2}\cdot 2(x-4)-(x-4)^2\cdot 2 \cdot \frac{1}{2}\cdot \left({\frac{1}{\sqrt{x}}}\right)}{4x}}

    The common denominator in the numerator is sqrt{x}. You'll get:
    \displaystyle{\frac{\frac{4x(x-4)-(x-4)^2}{\sqrt{x}}}{4x}}Expand the numerator and put together both denominators:

    \displaystyle{\frac{4x^2-16x-x^2+8x-16}{4x \cdot \sqrt{x}}} which is exactly the given result

    EB
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  4. #4
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    Thanks again guys!

    Looking at yours I realised I did make a stupid error of originally not expanding correctly.

    In an earlier part of the question, you were to expand without the denominator - there, 2x in the denominator instead of 4x!

    Thanks!
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