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Math Help - Differentiation .. Quadratic Rule problem :(

  1. #1
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    Differentiation .. Quadratic Rule problem :(

    hey guys i have a little problem i cant get my head around

    i undertsand that

    y=uv'+vu'

    but this question

    (x-9)(x-3)^{0.5}

    thats (x-3)^0.5 which is root (i dont know how to root in the maths tags

    but the answer i keep getting is

    (x-9)/(x-3)^{0.5} + (x-3)^{0.5}

    the real answer is (3x-15)/(2(x-3)^{0.5})
    Last edited by mr fantastic; November 24th 2008 at 02:59 AM. Reason: Fixed the latex typesetting errors in powers
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  2. #2
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    Quote Originally Posted by jimzer View Post
    hey guys i have a little problem i cant get my head around

    i undertsand that

    y=uv'+vu'

    but this question

    (x-9)(x-3)^{0.5}

    thats (x-3)^0.5 which is root (i dont know how to root in the maths tags

    but the answer i keep getting is

    (x-9)/{\color{red}2}(x-3)^{0.5} + (x-3)^{0.5}

    the real answer is 3x-15/(2(x-3)^0.5)
    The red 1/2 I added is needed. So you have \frac{x-9}{{\color{red}2}(x-3)^{0.5}} + (x-3)^{0.5}, which is the same as

    \frac{x-9}{2\sqrt{x-3}} + \sqrt{x-3} = \frac{x-9}{2\sqrt{x-3}} + \frac{2(x-3)}{2\sqrt{x-3}} = \frac{x-9 + 2(x-3)}{2\sqrt{x-3}} = \, ....
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  3. #3
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    Quote Originally Posted by jimzer View Post
    hey guys i have a little problem i cant get my head around

    i undertsand that

    y=uv'+vu'

    but this question

    (x-9)(x-3)^{0.5}

    thats (x-3)^0.5 which is root (i dont know how to root in the maths tags

    but the answer i keep getting is

    (x-9)/(x-3)^{0.5} + (x-3)^{0.5}

    the real answer is 3x-15/(2(x-3)^0.5)
    hey mate,

    as you stated you have an equation of the form

    y = u(x)v(x) for which you want to find the derivative which as you stated is solved using the product rule,
    i.e.,
    y'(x) =u'(x)v(x) + u(x)v'(x)

    here,
    (i) u(x) = (x-9) --> u'(x) = 1
    (ii) v(x) = (x-3)^(1/2) --> v'(x) = (1/2)*(x-3)^(-1/2) (I'm assuming you know the chain rule - please advise me if you dont and I'll be more than happy to explain)

    Therefore

    y'(x) = (1)*(x-3)^(1/2) + (x-9)*(1/2)*(x-3)^(-1/2) = (2(x-3) + (x-9))/(2*(x-3)^(1/2))
    = (1/2)*(3x - 15)/(x-3)^(1/2)
    Hope this helps,
    Let me know if you require any further assistance,

    David
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    thanks so much again youve helped me so much ..

    props to you if i get a sweet mark on the exam

    but 1 question. why is the 2 needed?
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    thanks a lot for that david its becoming clear
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  6. #6
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    Quote Originally Posted by jimzer View Post
    thanks so much again youve helped me so much ..

    props to you if i get a sweet mark on the exam

    but 1 question. why is the 2 needed?
    The derivative of (x - 3)^{1/2} is \frac{1}{2} (x - 3)^{1/2 - 1} (1) = \frac{1}{2} (x - 3)^{-1/2} = \frac{1}{2 (x - 3)^{1/2}}.
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  7. #7
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    Quote Originally Posted by jimzer View Post
    thanks a lot for that david its becoming clear
    hey mate,

    not a problem - just in reference to the (1/2) component I believe you were having troubles with, as you would know it composed in the derivative of,

    v(x) = (x-3)^(1/2)

    Here we employ the Chain rule, i.e. let w = x-3 then
    v = w^(1/2) --> v'(w) = (1/2)w^(-1/2) (i.e. d/dx (x^n) = nx^(n-1) )
    and w'(x) = 1

    Therefore by the Chain Rule,

    v'(x) = v'(w)*w'(x) or v'(x) = (dv/dw)*(dw/dx)
    Thus,
    v'(x) = (1/2)w^(-1/2) * (1)

    Recall w = x-3

    Therefore
    v'(x) = (1/2)(x-3)^(-1/2)

    Hope this helps,

    Regards,

    David

    ps - are you good with your logarithms? there is another technique which makes alot of these nasty derivates simpler to handle.
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  8. #8
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    thanks a lot mr fantastic and david, you guys have been of so much help

    and i'm okay with my logs but i rather not delve into them right this moment! might get a headache

    take care guys, thanks so much for all the help, it is greatly appreciated
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