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Derivative and Integral question

  1. Dec 16, 2007 #1

    asif zaidi

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    I have two problem relating to integrals and derivatives.

    Integrals: I can do but I am confused about the limit
    Derivative: I am having problem. Need some help with that. I have a solution but I think it is wrong.

    For each problem I have posted step by step what I am doing. Hopefully it will be easier in answering.


    Problem 1: Integral

    What is [tex]\int e ^{-t}[/tex] (limit is from 0 (on bottom) to 1 (on top))

    Problem 1: My solution
    My issue is step 6

    1. Take u = -t
    2. Use Chain Rule: dy/dx = (dy/du)(du/dx)
    3. du/dt = -1. Therefor -du = dt
    4. Find limits:
    For t=1: u= -1
    For t=0: u= 0
    5. New integral becomes (-1) [tex]\int e ^{u} du [/tex].
    6. My question/issue is about the new limits:
    Do I have to evaluate the integral with u=-1 on bottom of integral and u=0 on top of integral or the other way.



    Problem 2 Statement:

    y= [tex]x^{x^x}[/tex]. Find the minimum of y.

    Problem 2 Solution
    I am not really sure that I am doing this right. So plz advise

    I know how to find the derivative of [tex] x^x [/tex].

    So here are my steps

    1. make a = [tex] x^x [/tex] (I am taking the first two x's)
    2. Therefore y = [tex] a^x [/tex]
    3. dy/dx = (dy/da)(da/dx)
    4. Find dy/da:
    y = [tex] a^x [/tex] ​
    = [tex] e^{xlna} [/tex] ​
    dy/da = [tex] a^x [/tex] d(xlna)/da = [tex] (a^x) [/tex] [tex] x^{1-x} [/tex] ​
    5. Find da/dx
    a = [tex] x^x [/tex] = [tex] e^{xlnx} [/tex]​
    By productrule da/dx = [tex] x^x[/tex] (ln x + 1)​

    6. Multiple from step 4,5 and equate to 0.

    My question is: is step 4 right? Actually now that I am looking at it, I am wondering if step 5 is right also. Plz advise on this.



    Thanks

    Asif
     
    asif zaidi, Dec 16, 2007
  2. jcsd
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  3. Dec 16, 2007 #2

    CompuChip

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    Homework Helper

    For the first problem: you have taken u = -t. Plugging in the lower boundary for t gives you the new lower boundary for u (and similarly for the upper). So the new integral runs from u = -0 below to u = -1 above. You are allowed to interchange the boundary values though, as long as you put in a minus sign, that is:
    [tex]\int_a^b f(x) dx = - \int_b^a f(x) dx[/tex]
    In this case, it will cancel the minus you get from the substitution du -> -dt.
     
    CompuChip, Dec 16, 2007
  4. Dec 16, 2007 #3

    malty

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    Gold Member

    In your first problem, the limits were from t=0 to t=1

    When you substituted for u, and integrated with respect to du, you're limits were still t=o and t=1, if you were to swap these to be from t=1 to t=0 zero you would have to put a minus sign before the entire integral.
    remember [tex] \int_{a}^{b}f(x)dx=-\int_{b}^{a}f(x)dx[/tex]

    So you cannot put u=-1 on the bottom without altering the sign of the integral, because you cannot put t=1 on the bottom without changin sign.

    //Ah I see, compuchip already said this:smile:

    For part 2, you could save yourself a lot of hassle by just letting

    [tex] y=x^{x^x} =y=x^{x^2}[/tex] ;)
     
    Last edited: Dec 16, 2007
    malty, Dec 16, 2007
  5. Dec 16, 2007 #4

    ozymandias

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    Problem 1:
    You should evaluate your "new" integral between the "old" limits, expressed in terms of the "new" variable u.
    For example: the limit t=0 becomes u(0) = -0 = 0.
    The limit t=1 becomes u(1) = -1.

    Problem 2:
    The chain rule works on functions of functions. In your case, you took y=a^x, while you should've taken y=x^a (i.e. the ARGUMENT of the function should be a). The final result should have more lns in it ;).

    -----
    Assaf
    "www.physicallyincorrect.com"[/URL]
     
    Last edited by a moderator: Apr 23, 2017
    ozymandias, Dec 16, 2007
  6. Dec 18, 2007 #5

    asif zaidi

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    I came back to this derivative question after some time. Thanks for the response on integrals

    So the problem statement was to find the minimum of [tex] x^{x^x} [/tex]

    Problem solution

    As ozymandis said I was taking the wrong approach to find derivative. Here is the proper approach.

    Step1:
    [tex] x^{x^x} [/tex] = [tex] e^{e^{xlnx} ln x} [/tex]

    Step2: take derivative of above equation on right hand side

    y' = [tex] x^{x^x} [/tex] * derivative of ([tex] e^{e^{xlnx} ln x} [/tex])

    = [tex] x^{x^x} [/tex] * ( [tex] (x^x /x) [/tex] + ln x [tex] (x^x)' [/tex] )

    = [tex] x^{x^x} [/tex] * ( [tex] (x^x /x) [/tex] + ln x ([tex] x^x [/tex] )(ln x + 1) ) --- eq1

    Step3: Set eq 1 to 0 to find min

    [tex] x^{x^x} [/tex] * ( [tex] (x^x /x) [/tex] + ln x ([tex] x^x [/tex] )(ln x + 1) ) = 0

    I can simplify this to

    lnx (lnx + 1) + 1/x = 0 -----eq2

    Question

    1- How the heck do I evaluate this eq2. I ran this on matlab and the minimum value I get for x = 1. I never get to 0. Have I calculated the derivative properly.

    2. I ran the x^x^x on matlab and came up with a val of 0.832 so thats the value I should come up with once I evaluate eq2 but I cannot get past above problem.

    What am I doing wrong?


    Thanks

    Asif
     
    asif zaidi, Dec 18, 2007
  7. Dec 19, 2007 #6

    ozymandias

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    Hi Asif,

    Take q = ln(x), then solve the quadratic equation.
    Edit: whoops, just saw the 1/x. :)
    I don't believe this equation can be solved symbolically.
    Edit 2: I just re-differentiated, your result still seems in error.
    Try:
    [tex]y(x) = x^{x^x} = x^a[/tex] where [tex]a=x^x[/tex] and use the chain rule.

    Assaf
    "www.physicallyincorrect.com"[/URL]
     
    Last edited by a moderator: Apr 23, 2017
    ozymandias, Dec 19, 2007
  8. Dec 19, 2007 #7

    asif zaidi

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    Hi Ozymandias:

    Seems I am doing something wrong.
    Actually I tried your method of replacing a=[tex] x^x[/tex] and I still got the same equation that I had before - namely the 1/x component.

    Here are the steps - I cannot figure out what I am doing wrong.

    Step1:
    y = [tex] x^a [/tex] = [tex] exp ^{alnx} [/tex] where a = [tex]x^x[/tex] = [tex] exp ^{xlnx} [/tex]


    Step2
    y' = [tex]x^{x^x} [/tex] (a'lnx + a/x) <--- 2nd component from chain rule. As you can see I still get the 1/x component.

    Step3
    a' = [tex] x^{x^x}[/tex] (1+ lnx)

    Step4
    Substituting from step3 into step2
    y' = [tex]x^{x^x} [/tex] ( [tex] x^x[/tex] lnx (lnx + 1) + ([tex]x^x [/tex])/x )
    = ([tex]x^{x^x} [/tex])([tex] x^x[/tex]) ( 1/x + lnx(lnx+1) )

    Step5:To find min
    0 = [/tex])([tex] x^x[/tex]) ( 1/x + lnx(lnx+1) )

    Which is where I was. So obviously I am doing something wrong. Step2. I have re-read the text book and I think I have understood it properly. I did the other problems and got the right ans.

    This is driving me crazy.

    Thanks

    Asif
     
    asif zaidi, Dec 19, 2007
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