x^y=y^x

Discussion in 'Physics & Math' started by Jerrek, Mar 25, 2003.

  1. Jerrek Registered Senior Member

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    How would you go about differentiating this question, say with respect to x? Find y'. This is definitely giving me some issues.

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  3. Persol I am the great and mighty Zo. Registered Senior Member

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    Damn... tough one

    x^y=y^x
    y=x^(y/x)

    by analysis this is true if y/x=1 (although I'm not sure how to prove it besides plotting a contour of x^y-y^x=z)

    so this would simplify to x=y

    x'=1
     
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  5. lethe Registered Senior Member

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    logarithmic differentiation. not too bad. i get dy/dx = [ln y - y/x]/[ln x - x/y]
     
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  7. Jerrek Registered Senior Member

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    But lethe, I did that, but you still end up with a y in the other side. Is that acceptable? (Pardon me, I'm second year and I'm not fond of Calculus.)
     
  8. On Radioactive Waves lost in the continuum Registered Senior Member

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    is that a homework problem? that seems damn tough
     
  9. lethe Registered Senior Member

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    x^y = y^x is a transendental equation, so i would expect there to be a y on both sides of the equation. if you can t solve the original equation for y, i don t see why you should be able to solve it in the derivative equation.
     
  10. Jerrek Registered Senior Member

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    Thank you. No it isn't a homework problem, just something one of my friends threw at me to make me think.
     
  11. RKBA Registered Member

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    For what it's worth, here's what Maple thinks about it:

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  12. HallsofIvy Registered Senior Member

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    To answer the question that was asked earlier, yes, it quite acceptable to write dy/dx in terms of both x and y, in fact it might be NECESSARY. Of course, theoretically, one should be able to solve for y as a function of x and replace it but the whole point of "implicit differentiation" is that it is applicable when you CAN'T solve for x.

    Implicit differentiation is applicable even when y is NOT a function of x. A simple example is : x<sup>2</sup>+ y<sup>2</sup>= r<sup>2</sup>. While it is easy to solve for y, it's double valued: we have to choose either + or -. On the other hand,
    differentiating the equation gives 2x+ 2yy'= 0 so y'= -x/y.

    Here it wouldn't make sense to ask for the derivative at a specific value of x: there may be two different values of y for a single value of x- you have to give both x and y.
     

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