Graph of x^y=y^x

graph of x^y=y^x,
check out how to find the parametric equations: • Solutions to x^y=y^x
how to graph x^y=y^x, • Graph of x^y=y^x
derivative of x^y=y^x using implicit differentiation • derivative of x^y=y^x,...
derivative of x^y=y^x using total differential (calculus 3), • derivative of x^y=y^x
www.blackpenredpen.com
math for fun

[@royler8848]

Its 3 AM,but I cannot sleep without seeing the graph of this thing xD

[@KaiCyreus]

"The Ellen Teeth Power"

[@pixelninja5766]

please do a video on x^x=y^y./ I dont un derstand that graph

[@adamkirsopp492]

This becomes so much easier when you accept that 0^0=1

[@marcelweber7813]

I still don't understand why we can assume y=tx in the first place. It seems like a logic step, but is it proven that there aren't more solutions?

[@JCCyC]

Looks like a hyperbola but ain't.

[@tylerhuang7507]

Why don't the negatives work?

[@spelunkerd]

I think I'm missing something basic. If x is plotted on the x axis, and y is plotted on the Y axis, then you get the diagonal curve y=x. Are you plotting t on the z axis?

[@nathanielkilmer5022]

I can say unequivocally that 0^0=0^0.

[@manawer720]

I have a pretty heird question about that thing of "0^0 = 0^0", it's "ok" to say that "1/0 = 1/0"? I know that it may not exist something like "it's ok" sort of thing here, but it made me think a lot

[@DarkMage2k]

"Differential symbols make me happy"

[@Harlequin314159]

0:50

[@DoesMahBlockLookBig]

Off topic, but I tried seeing if integrating (e^sinx - e^-sinx)/2 was possible. Couldn’t find a way, so I decided to plug the function into desmos, and weird enough, it looks exactly like a sin function. In fact, it looks EXACTLY similar to 1.175 sin x. There are very VERY small differences between f(x) = (e^sinx - e^-sinx)/2 and f(x) = 1.175 sin x. Each function had the same maxima and minima, and also the same x-intercepts.

[@macronencer]

I was hoping for some great content from this one, after seeing part one - and you delivered! I had to stop and rewind a few times to make sure I fully understood what you had done, but I got it in the end. Thank you.

[@SocksWithSandals]

I was thinking about this in my head all week.

[@danuttall]

1:50

[@albertosoria6564]

You're a really genius in maths. It's incredible how can you know all of that XD.

[@Galileo2pi]

Keep it up mate, I love it.

[@iabervon]

It's worth substituting t=1/u. You'll find that when the parameter is greater than 1, the x and y values are the y and x values for 1/t. So you don't need to work out that region even if you're interested in where the curve is for values of the parameter.

[@rubensenouf1813]

You're amazing ! Thanks for your videos !