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Math Help - Exponent/Log Functions

  1. #1
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    Post Exponent/Log Functions

    You told me to post a couple more on here to make sure I had gotten this...
    This one is going to prove more difficult for me.

    f(x) = the 4th root of (2x+1/1-3x)....the whole fraction under the 4th root.

    u(x) = the 4th root of u or u^(-1/4)
    u'(x) = -1/4u^(-5/4)
    v(x) = [2x+1/1-3x]
    v'(x) = (1-3x)(2) - (2x+1)(-3)/(1-3x)^2
    v'(x) = (2-6x) - (-6x-3)/(1-3x)^2
    v'(x) = -1/(1-3x)^2

    If that is correct, I will move on with
    u(x) * v'(x) - v'(x) * u(x)/u(x)^2
    Last edited by becky; May 6th 2007 at 11:55 AM. Reason: misspelling
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  2. #2
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    Post I must not be getting this....

    Wasnt asking for an answer, just if the steps were correct so far.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by becky View Post
    Wasnt asking for an answer, just if the steps were correct so far.
    the fourthroot of u is not u^(-1/4), so the first one is wrong. it's just u^(1/4)

    the second one is incorrect also. you made a mistake when expanding the brackets. you will have in the top 2 - - 3 = 2 + 3 = 5, you forgot that there was a double negative
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by becky View Post
    If that is correct, I will move on with
    u(x) * v'(x) - v'(x) * u(x)/u(x)^2
    what's this for? shouldn't you be doing the chain rule? you have u(v(x)) so

    [u(v(x))]' = u'(v(x))*v'(x)
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  5. #5
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    Re;

    Re:
    Attached Thumbnails Attached Thumbnails Exponent/Log Functions-dervative-2.jpg  
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