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Homework Help: Few Problems (Derivatives)

  1. Nov 13, 2006 #1
    I need a little help on the following problems:

    Find the 2nd derivative at the indicated point:
    y= x(ln x) At the point (1,0)

    I got to:

    y''= [tex]
    \frac{1}{x}
    [/tex]

    So do I just sub in 1 for x? And then the answer would be 1?

    Also, I need help on the following problem:

    Evaluate the first derivative at the given value X: x=0

    [4^(3x)(x^3 - x + 1)^(1/5)] (And then all of that raised to the -2)
    -------------------------
    (x^2 + x + 1)^(4)
     
  2. jcsd
  3. Nov 13, 2006 #2

    JasonRox

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    Homework Helper
    Gold Member

    Yeah, basically put 1 in the equation for the first question.

    For the second one, I see no trick. I guess you have to painfully use the Quotient Rule, Chain Rule and Product in the right order. Have fun doing that. :S

    Actually, I'd probably do it using first principles. It seems like the easier way out. (First principles is finding the derivative using the limit definition.)
     
  4. Nov 14, 2006 #3

    HallsofIvy

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    Science Advisor

    JasonRox, you have a wicked sense of humor! Anyone trying to find the derivative of that second function using the limit of the difference quotient is guarenteed to go insane!

    TommyLF, I would recommend going ahead and incorporating that -2 exponent into the numerator and, because that is a negative exponent, writing the whole thing as a product of terms with negative exponents:
    [tex]\frac{[4^{3x}(x^3- x+ 1)^{1/5}]^{-2}}{(x^2+ x+ 1)^4}= 4^{-6x}(x^2-x+1)^{-2/5}(x^2+ x+ 1)^{-4}[/tex]
    and use the product rule and chain rule.
     
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