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Exponential rule on y = (1+x)^(1/x)

  1. Mar 11, 2012 #1
    1. The problem statement, all variables and given/known data


    I am of the conclusion that, under any circumstance, the extended exponential rule can not be applied to (1+x)[itex]^{1/x}[/itex].

    Thus, there is no way for the extended exponential rule to arise when taking the derivative of:

    f(x) = (1+x)[itex]^{\frac{1}{x}}e^{x}[/itex]

    For instance, if my first step for finding the derivative of this function was to apply the product rule, i'd get:

    f'x = ((1+x)[itex]^{\frac{1}{x}})'(e^{x})[/itex] + ((1+x)[itex]^{\frac{1}{x}})(e^{x})'[/itex]

    And in the next step, if I were to take the derivatives by first applying the exponential rule to (1+x)[itex]^{1/x}[/itex],

    I would get an incorrect outcome because while I could apply the exponential rule to something like b^x, I would not be able to apply exponential rule to something like (b+x)^x

    Is this correct so far?
     
  2. jcsd
  3. Mar 11, 2012 #2

    Fredrik

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    If what you mean by the "exponential rule" is the rule for how to take the derivative of something like ax, then you're right that you can't use it for things like xx. It is however easy to rewrite the thing that you don't know how to deal with, as something that you do know how to deal with. For example,
    $$x^x=e^{\log x^x}=e^{x\log x}.$$
     
  4. Mar 11, 2012 #3

    SammyS

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    As Fredrik said, you can often change the way you an expression to make it easier to work with.

    For the example you give, you can use logarithmic differentiation, or rewrite the expression as follows.
    [itex]\displaystyle
    f(x)=\Large e^{\ln\left((1+x)^{1/x}\right) } e^x[/itex]

    [itex]\displaystyle =\Large e^{\left( \frac{\ \ln(1+x)\ }{x} +x\right)}[/itex]
     
  5. Mar 11, 2012 #4
    Fredrik & Sammy, thank you both very much. I did not see this problem in the way you presented it. Thank you for showing me this way to approach functions of the form x[itex]^{x}[/itex].
     
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