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Thread: homework help needed with differentiation

  1. September 17th 2007, 01:16 PM #1
    rpatel
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    homework help needed with differentiation

    I have attached a question which i kind of confused with. I understand the differentiation parts but i don't understand the simplifying bits.

    The main simplification bit i don't understand is the 12x^2 -4x bit which i have circled in the attached image. I think that the 12x^2 -4x bit comes from the simplifying (squareroot 3x-1) * 2x but i not sure and if it does i don't understand this simplification on how it was done.

    I would greatly appreichate any help in solving this problem.

    thanks
    Attached Thumbnails Attached Thumbnails homework help needed with differentiation-mathsq.jpg  
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  2. September 17th 2007, 06:03 PM #2
    Jhevon
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    Quote Originally Posted by rpatel View Post
    I have attached a question which i kind of confused with. I understand the differentiation parts but i don't understand the simplifying bits.

    The main simplification bit i don't understand is the 12x^2 -4x bit which i have circled in the attached image. I think that the 12x^2 -4x bit comes from the simplifying (squareroot 3x-1) * 2x but i not sure and if it does i don't understand this simplification on how it was done.

    I would greatly appreichate any help in solving this problem.

    thanks
    we have f'(x) = x^2 \cdot \frac 32 (3x - 1)^{- \frac 12} + 2x \sqrt {3x - 1}

    = \frac {3x^2}{2 \sqrt {3x - 1}} + 2x \sqrt {3x - 1}

    the LCD here is 2 \sqrt {3x - 1}, so we get:

    f'(x) = \frac {3x^2 + 2 \cdot 2x \left( \sqrt {3x - 1} \right)^2}{2 \sqrt {3x - 1}}

    = \frac {3x^2 + 4x(3x - 1)}{2 \sqrt {3x - 1}}

    = \frac {3x^2 + 12x^2 - 4x}{2 \sqrt {3x - 1}}

    ...
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  3. September 17th 2007, 06:49 PM #3
    Krizalid
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    Quote Originally Posted by rpatel View Post
    Let's set y=f(x)=x^2\sqrt{3x-1}\implies y^2=x^4(3x-1)=3x^5-x^4

    Let's take now the derivative

    2yy'=15x^4-4x^3\implies y'=\frac{x^3(15x-4)}{2y}

    Back substitute and we happily get that f'(x)=\frac{x^3(15x-4)}{2x^2\sqrt{3x-1}}=\frac{x(15x-4)}{2\sqrt{3x-1}}

    Does that make sense?
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