Differentiate y=x^y

Snotzenfartz

What's the derivative of [latex]y = x^y[/latex] ?

I tried implicit differentiation and got:

[latex]y' = y'yx^{(y-1)}[/latex]

which doesn't make sense unless y' = 0 or infinity.

Snotzenfartz

Taking logs and differentiating i get:

[latex]logy = ylogx[/latex]

[latex]y'/y = y/x + y'logx[/latex]

Solving for y' gives:

[latex]y' = y^2 / (x - xylogx)[/latex]

Is this right?

Yakuza

How can something be a function of itself? Of course, after what happened with the "sum to infinity of 1+2+4+8... being equal to -1" thread, I'm prepared to be corrected, but a function (not equation) whose output variable (I can't remember the fancy mathsy term for this so I've substituted a computer science one) exists on the LHS and RHS makes no sense to me at all, as you get into recursion etc. Here, there be dragons!

Edit : The "mathsy" term i was looking for was "dependent variable"

Snotzenfartz

I've been trying to differentiate the function,

[latex]y=x^{x^{x^{x^{.^{.^{.}}}}}}[/latex]

If you raise x to the power of each side it obviously doesn't change the function since there are infinitely many x's already. So we can implicitly define the function as,

[latex]y=x^y[/latex]

It's a bit sneaky but nothing paradoxical.

Yakuza

Barring the special cases of x=0 and x=1, anything else raised to itself (infinitely many times) is pretty much going to snap right to zero (for 0<abs(x)<1) or infinity (abs(x)>1) (straight away, rendering a derivative (in terms of a rate of change or the slope of the line at that point) fairly meaningless, no?

Splagmata

Nah, you're thinking of,

[latex]y = ((((((x^x)^x)^x)^.)^.)^.[/latex]

Which would have the implicit definition,

[latex]y=y^x[/latex]