Derivative of infinite x^x

[Luciferr]

How would you solve this derivative of doom?

 

[Deleted member 4993]

How would you solve this derivative of doom?
View attachment 32792

try

y = x^y

 

[Steven G]

@Subhotosh Khan,

\(\displaystyle If\ y=x^y,\ then\ y = x^{x^{x^{x^...}}}.\)
Isn't that true?

 

[Cubist]

@Subhotosh Khan,

\(\displaystyle If\ y=x^y,\ then\ y = x^{x^{x^{x^...}}}.\)
Isn't that true?

I agree, but @Subhotosh Khan had a great idea there, and this must be the necessary adjustment... [imath]y=(x^x)^y[/imath]

 

[Steven G]

I agree, but @Subhotosh Khan had a great idea there, and this must be the necessary adjustment... [imath]y=(x^x)^y[/imath]

Thanks. I didn't see the adjustment [imath]y=(x^x)^y[/imath]. I need to have my vision checked.

 

[topsquark]

Thanks. I didn't see the adjustment [imath]y=(x^x)^y[/imath]. I need to have my vision checked.

To the corner with you, you scurvy swine! (I could use someone to play rock paper scissors with.)

-Dan

 

[Steven G]

To the corner with you, you scurvy swine! (I could use someone to play rock paper scissors with.)

-Dan

Why are you in the corner? What did you do this time around?

 

[topsquark]

Why are you in the corner? What did you do this time around?

I advised someone to use the Lambert W function where it couldn't actually be applied.

I'm a baaaad boy.

-Dan

 

[Steven G]

I advised someone to use the Lambert W function where it couldn't actually be applied.

I'm a baaaad boy.

-Dan

Sloppy, very sloppy.

 

[Deleted member 4993]

What did you do this time

Blasphemy!!!!

He invoked Lambert without any basis.......

 

[topsquark]

Blasphemy!!!!

He invoked Lambert without any basis.......

Help! I'm being lambasted about Lambert!

-Dan

 

[Steven G]

Help! I'm being lambasted about Lambert!

-Dan

What do you need help with? What have you tried? Did you read the posting guideline?
Please start a new thread!

 

[Cubist]

Here's the graph of [imath]y=(x^x)^y[/imath]



The graph starts near, or on, (0, 1) depending on how you view 0^0. It goes through (1,1) and then appears to double back where y=e. Then there's a vertical asymptote at x=1.

And there should be a point at (-1, -1)

 

[Steven G]

Here's the graph of [imath]y=(x^x)^y[/imath]

View attachment 32807

The graph starts near, or on, (0, 1) depending on how you view 0^0. It goes through (1,1) and then appears to double back where y=e. Then there's a vertical asymptote at x=1.

And there should be a point at (-1, -1)

Thanks for this graph. Now this is initially surprising. Can you show a graph of the derivative?

 

[BigBeachBanana]

Here's the graph of [imath]y=(x^x)^y[/imath]

View attachment 32807

The graph starts near, or on, (0, 1) depending on how you view 0^0. It goes through (1,1) and then appears to double back where y=e. Then there's a vertical asymptote at x=1.

And there should be a point at (-1, -1)

I used Desmos and it gave me something different for [imath]y=(x^x)^y[/imath]


This is what I got for the derivative.

 

[Cubist]

I used Desmos and it gave me something different for [imath]y=(x^x)^y[/imath]
View attachment 32809

This is what I got for the derivative.
View attachment 32810

I used Desmos too, but I zoomed out the y axis scale to see how it folds back for higher y values!
Example x = 1.3 implies y ≈ 4.22484229404955304071 or y ≈ 1.93426050277962582024

 

[Cubist]

Here's a graph of the gradient (red and green curves) with the original shown in blue. Note that the graph is cut at the thick black line. Beneath that line the y axis is on a different scale, since the gradient is very steep and negative. The x axis values are aligned in both sections of the graph.

Note: The gradient is exactly 1 at (1, 1)

EDIT: This only shows the gradient at the (x,y) points where the equation [imath]y=(x^x)^y[/imath] is true, even though the expression I obtained for the gradient gives a field of values (and this is probably what is shown in post#15)