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Derivative of e^(x^x) with respect to x

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Derivative of e^(x^x) with respect to x

    2. Relevant equations



    3. The attempt at a solution

    Computed using wolframalpha. I have attached the image . Would anyone explain to me the part I have highlighted with the blue box.
     

    Attached Files:

  2. jcsd
  3. Aug 5, 2011 #2

    BruceW

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    imagine that at first, you don't realise that both of the variables are x, so then you have an equation like this: [itex]u^v[/itex] So then, he calculates the total derivative of this equation in a general way, then at the end he replaces u by x and v by x.

    Edit: ha, I'm talking about wolfram as a 'he'. Also, there is another way to solve this.
     
    Last edited: Aug 5, 2011
  4. Aug 5, 2011 #3

    tiny-tim

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    hi labinojha! :smile:

    this is the partial derivative version of the https://www.physicsforums.com/library.php?do=view_item&itemid=353"

    if [itex]a[/itex] depends on [itex]b_1,\cdots b_n[/itex], and [itex]b_1,\cdots b_n[/itex] depend only on [itex]c[/itex], then:

    [tex]\frac{da}{dc}\ =\ \frac{\partial a}{\partial b_1}\frac{db_1}{dc}\ +\ \cdots \frac{\partial a}{\partial b_n}\frac{db_n}{dc}\ =\ (\mathbf{\nabla_b}\,a)\cdot \frac{d\mathbf{b}}{dc}[/tex]​

    in your case, a is xx, b1 and b2 are u and v,

    and so a is a function a(u,v) of two variables, and we need to apply the chain rule to each variable separately

    (btw, easier would be to say xx = exln(x) :wink:)
     
    Last edited by a moderator: Apr 26, 2017
  5. Aug 5, 2011 #4
    [itex]\frac{\partial}{\partial x}[/itex][itex]u^{v}[/itex]=[itex]\frac{\partial}{\partial x}[/itex][itex]u^{v}[/itex].[itex]\frac{\partial}{\partial x}[/itex][itex]u+\frac{\partial}{\partial x}[/itex][itex]u^{v}[/itex].[itex]\frac{\partial}{\partial x}[/itex]v

    Can i get any reason or possibly a derivation for this ?

    Hi tiny-tim!
    I had been writing so your answer came before I questioned. :)

    http://www.ucl.ac.uk/Mathematics/geomath/level2/pdiff/pd10.html
     
    Last edited: Aug 5, 2011
  6. Aug 5, 2011 #5
    Hi Bruce!
    The other way i used to do it was to take the natural logarithm of both the sides(one side of the equation being y to suppose it as the function) two times in a row and then differentiating them both sides .

    Is this what you were talking about ? :)
     
    Last edited: Aug 5, 2011
  7. Aug 5, 2011 #6

    BruceW

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    I was talking about the thing tiny-tim said:
    [tex] x^x = e^{xln(x)} [/tex]
    But yes, the other way you do it is right as well. I guess there are several equivalent ways to do this problem.
     
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