# x^y=y^x

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

1. ### JerrekRegistered 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.

3. ### PersolI 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

5. ### letheRegistered Senior Member

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logarithmic differentiation. not too bad. i get dy/dx = [ln y - y/x]/[ln x - x/y]

7. ### JerrekRegistered Senior Member

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1,548
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 Waveslost in the continuumRegistered Senior Member

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is that a homework problem? that seems damn tough

9. ### letheRegistered 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. ### JerrekRegistered Senior Member

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1,548
Thank you. No it isn't a homework problem, just something one of my friends threw at me to make me think.

11. ### RKBARegistered Member

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

12. ### HallsofIvyRegistered 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.