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Homework Help: Deriatives problem

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data

    find y=d/dx(e^x.e^x^2.e^x^3.................e^x^n) at 0<=x<=1
    at x=0.5
    2. Relevant equations

    3. The attempt at a solution

    so y=e^(x+x2+x3....xn)
    1-i tried taking natural log on both sides ,thn differentiate.
    it does not help..neither does using the d/dx(uv) help....both gives
    y= e(summation of i=1 to n of x^r).(1+2x+3x^2.....nx^(n-1) )
    = e^(x(x^n-1)/x-1).summation of nx^(n-1).
    i cant simplify thies any further.and it looks horrible when x=0.5

    and does not give me the slightest feeling that its the correct solution.. the answer is (4e)..but i have been told to try till i reach the solution!!!!
  2. jcsd
  3. Sep 17, 2009 #2
    and the answer also asks for a condtion.
  4. Sep 17, 2009 #3


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    Science Advisor

    I don't see why writing it as [itex]y= e^{x+x^2+x^2+\cdot\cdot\cdot+x^n}[/itex] wouldn't work. The derivative of that is, of course, [itex](1+ 2x+ \cdot\cdot\cdot+ nx^{n-1})e^{x+x^2+ \cdot\cdot\cdot+ x^n}[/itex] and that is to be evaluated at x= 1/2.

    [itex]x+ x^2+ \cdot\cdot\cdot+ x^n[/itex] is a geometric series except that it is missing the initial "1". Its sum is given by [itex](1- x^{n+1})/(1- x)[/itex]. And the derivative, term by term is just the derivative of that: [itex]1+ 2x+ \cdot\cdot\cdot+ nx^{n-1}}= [(n+1)x^n(1- x)+(1- x^{n+1})]/(1- x)^2[/itex]. Both can be evaluated at x= 1/2.

    The only difficulty appears to be that the result should be 4e for all n. It certainly is not. For example when n= 1, this is just [itex]e^x[/itex] which has derivative [itex]e^x[/itex]- evaluated at x= 1/2 that is [itex]e^{1/2}[/itex]. For n= 2, this is [itex]e^{x+x^2}[/itex] and its derivative is [itex](1+2x)e^{x+x^2}[/itex] which is [itex]2e^{3/4}[/itex] when x= 1/2.

    Are you sure you have stated the problem correctly? Is it not, "find the limit, as n goes to infinity, of the derivative of [itex]e^{x+ x^2+ \cdot\cdot\cdot+ x^n}[/itex]"? That would be, not, 4e, but [itex]4e^2[/itex].
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