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Dy/dx exponential function

  1. Dec 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Find dy/dx.
    [itex]y=x^{lnx}[/itex]


    2. Relevant equations



    3. The attempt at a solution
    [itex]ln y= ln x^{ln x}[/itex]
    [itex]ln y= (ln x)(ln x)[/itex]

    Taking the derivative now:

    [itex]\frac{1}{y}y'= (\frac{1}{x})(\frac{1}{x})[/itex]
    [itex]\frac{1}{y}y'= (\frac{1}{x})^2[/itex]

    Multiply by y:

    [itex]y'= (\frac{x^{lnx}}{x^2})[/itex]

    But it's not the right answer :?
     
  2. jcsd
  3. Dec 1, 2011 #2

    Mark44

    Staff: Mentor

    Are you trying to say that d/dx(fg) = f'g'?
     
  4. Dec 1, 2011 #3
    I'll give you a hint for the way I would solve this problem. Use the chain rule!
     
  5. Dec 1, 2011 #4
    Wow forget the basics and you completely screw up lmao..

    So then I get:

    [itex]y'=(\frac{lnx}{x}+\frac{lnx}{x})(y)[/itex]
    [itex]y'=(\frac{2lnx}{x})(x^{lnx})[/itex]
    [itex]y'=\frac{2x^{lnx}lnx}{x}[/itex]
     
  6. Dec 1, 2011 #5
    Too bad the chain rule doesn't work here.
     
  7. Dec 1, 2011 #6

    dextercioby

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    Looks ok now.
     
  8. Dec 1, 2011 #7
    anything is possible

    zDzwZ.jpg
     
  9. Dec 1, 2011 #8
    Not the answer the book gives..
     
  10. Dec 1, 2011 #9
    Yeah. I know. It is a different way of solving the problem, but the answer is still right.
     
  11. Dec 1, 2011 #10

    dextercioby

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    Needn't have copied the fine (and smart) calculation. The answer is nonetheless correct.
     
  12. Dec 2, 2011 #11

    HallsofIvy

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    Staff Emeritus
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    Just to add: there are two serious mistakes one could make in differentiating
    [itex]f(x)^{g(x)}[/itex]:

    1) Treat g(x) as if it were a constant- [itex]g(x)f(x)^{g(x)- 1}[/itex]
    2) Treat f(x) as if it were a constant- [itex]f(x)^{g(x)}ln(g(x))[/itex]

    The peculiar thing is that the correct derivative is the sum of those two "mistakes".
     
  13. Dec 2, 2011 #12

    Curious3141

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    Homework Helper

    My preferred way to do this:

    [itex]y={(e^{\ln x})}^{\ln x} = e^{(\ln x)^2}[/itex]

    Differentiate with Chain Rule,

    [itex]y' = e^{(\ln x)^2}(\frac{2\ln x}{x}) = \frac{2x^{\ln x}}{x}[/itex]
     
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