Calculus: Derivative of x^(x^x)

[Carebearbull]

Yo dawg i herd you like x's, so I put a x to the power of x in yo power of x so you can derive while u derive while u derive.

- X to the Z - X(z)ibit.

where Z = X^x

[VechtMalthos]

if you don't replace (1/x)*(x^x) with x^(x-1), you can then factor out an x^x

if you do that and distribute the (ln x) the resulting equation is indeed 'elegant':

Dx[x^x^x] =

x^x^x * x^x ( ln^2 x + ln x + 1/x )

[nazierDK]

dude U e a hero :)

thanks a lot

[Darksp00n]

you are my hero .. thanks :)

[ThePimpinGrandpa]

@khanacademy One of my stupid friends from class showed my teacher this video and now she wants us to differentiate the long equation at 8:31... omg...

[jaymejamjar]

THANK YOU! calculus exam in two days and I was so stuck. I'm definitely going to need to keep watching a few more derivative videos before i get the hang of it, but this was so helpful. thank you!

[Roxas99Yami]

@gypsytearss1

than you just have to start derivating the complex numbers ... ln(-3) is actually a complex number which more precisely is (i*pi) + the integral from 1 to 3 of (1/x)dx ...

[lmq0209]

Awesome video! Thank you for posting!

[Dave67004]

Awesome! Nice little problem... I guess you could continue and differentiate

x^[x^(x^x)] using this result, and maybe determine a general result for when x appears n times in the function to be differentiated.

[Roxas99Yami]

@Dave67004

not that hard, no need to understand a rule from just 2 functions, you differentiate x^(nx) and ur done ^_^

[jensamis]

Dude, you are my hero. I giggled at the rediculousness of it all, but somehow completely understood it. Just lovely.

[qtzlctl2012]

hey! what's with calling hairy things "not that pleasant to look at"? :-p

[Scrapingthebottom]

Huh?



[sschinychin]

In Soviet Russia, you don't find the derivative. DERIVATIVE FINDS YOU!!!!!!!1

[vadimthemachine]

@sschinychin that joke is as old as the powdered milk from your mom's breasts.

[sschinychin]

@vadimthemachine

IKR

[dragooner4]

so the final derivative is x^x(ln x +1) so can u use distributive property and get

x^x*ln(x)+x^x

then use log rule and get ln(x^(x^x)) + x^x

[ffg9011]

Thanks for the tip of the ln y, greetings from Costa Rica.

[buhoman3]

you use the word hairy in almost every vid...

[learning112233]

how can brilliant mathematician be funny ?

[moosebop]

In all seriousness, you my friend are going to be the reason that myself, as well as many others, actually graduate from college. You were made for teaching.

[xxcowslayerxxx]

best proof i have seen on youtube, nobody else i could find actually explained the product rule part they just wrote it out, so thank you for helping me with my incompetence

[Mernusify]

just to say, the actual derivative is......

X^(X^X+X)(ln X)^2+X^(X^X+X)ln X+X^(X^X+X-1)

anyways not to confuse you.

[Cisco121490]

TY, great vid.

[5600981]

if I were to stumble upon this question without knowing derivative of x^x , do I use the logarithmic differentiation twice?

[Edgtheow]

@5600981 Yes, I believe so.

[Macybacy]

Nice one dude! Pretty simple and elegant solution to this problem!

[kurozaki45]

we still have the nasty x^x there I LOL'd so hard hahahaha

[dannystoll84]

Figured this out a month ago the hard way - by using y=e^(ln(x^x^x)) and differentiating from there… let's just say this is a lot more elegant

[nawptown]

great job. helped me out so much

[xXxWat2K7sonxXx]

This one is a classic and this very good anyway.

[mikechis101]

How do you know to take the ln of both sides? Is there some property that I don't know?

[liberate9]

@mikechis101 The fact that it is x ^ x gives it away. You will want to break it down into a problem without exponents since they are both variables. If one were constant you wouldn't need to break it down, but since they are, you have to so u can work with them individually.

[Yu2Kal]

fantastic problem, ever better solution

[dannyboy12357]

brilliant

[pipsticles01]

Great video, very easy to understand.

[sanjor8r]

is this in hd? looks better than the other vids

[fiercefirefly]

Thanks for posting these videos!

[yassin908]

wow! Ur superp, my friend always ask me this question but now i get it! ....all i can says is Thanks!

[ctw005]

DUUUUUUUUUUUUDE. Amazing. AAAAAAAMAAAAAAAAZING.

[Subaangen]

does he write with a mouse???

[gobberpooper]

he probably has a wacom tablet

[blackspoty2]

no i think a pen

[DJPirrone]

Great going on the new interface your videos look so much better

[inesescom]

is this the same thing as derivative of x^x^x? (with no "( )" around the upper part of the mess)

[inesescom]

i guess it is;)

[outofmylaboratory]

yeah it is because exponents come first in order of operations. He just put the parentheses so it was more clear.

[veteran25]

Ur MY HERO! I am learning more from your website than I have learned in all 3 months of AP Calculus BC

[213shahrzad]

thank youuu :]

[pive29]

what is the derivative of 3^(4x)??

[riku0228]

Thank you for helping me with such a problem. I appreciate your work

[woodenjaw]

YEAHHH!!! I got it right...

Thanks Sal!

[protosis100]

orale esta con ganas maestro. si pues tu si explicas mas faciles las matematicas. exelente

[duyu]

I can watch these videos in HD all day. Thanks for making new vids!

[carelessmorning]

Your writing with the mouse is great . :D

[SticksOnCrack]

he's using a tablet

[carelessmorning]

Ah, I see. His older videos seemed to have bad quality. Which is probably when he was just using mouse.

[gusdearaujo10]

He used a tablet, but the program was Paint with (which has no anti-aliasing, except in Windows 7).

[AUTubeN22]

oops sorry wrong class! :-( whats the name of the first class plz?

[rockYhre]

Great video! Especially using the HQ. I enjoy watching these videos a lot.

[ad2181]

Excellent.

[metainfinity]

Yeah, but what's the integral of x^x?

[calvinhobbesliker2]

That integral can't be done.

[metainfinity]

Why's that?

[calvinhobbesliker2]

It just doesn't. I tried to use an online integrator and it failed

[metainfinity]

Well yeah, I've also tried integrating it on a calculator. I'd like a little more reason than that though.

[calvinhobbesliker2]

It just doesn't have a simple formula for the function. It can be expressed as an infinite sum(using taylor series), but not in a finite way.

[Kosekans]

@calvinhobbesliker2 Sure, it can. The anti-derivative only can't be expressed in terms of elementary functions.

[DrJobinsp]

it looks so easy when you break it down in to smaller steps...

thats how math is ah :P

[Spoodily]

Youtube is the best thing to happen to learning math since numbers.

I feel confident that I can take any teacher's math class and pass their tests because Youtube has all the videos I would ever need to learn any material.

[vickiormindyb]

What does that ^ sign mean?

I now have absolute proof that a good teacher can be the difference in how the student learns.

In driving lessons, I switched my instructor and today's lesson was stress free.

The same with Math-and Chemistry. Without Sal doing these videos, I'd be failing Math and Chemistry.

[thelastwords]

I agree with that.. Sal is awesome!

the ^ symbol is the exponent symbol on the computer. Just like * means multiplication and / means division!

[lookwhoscomin]

hey sal i had an interview at a university and i was asked to firstly differentiate x^x and subsequently differentiate x^x^x^x^x... (infinitely many x's)

I could not solve the follow up question.

I did some research thereafter and found out that i should differentiate x^y implicitly If that's the case could you explain why?

[VeryEvilPettingZoo]

First ask, for fixed a, what does a^a^a^... mean? It means the limit of sequence b(1)=a, b(2)=a^a, b(3)=a^(a^a), etc.. If the limit exists, call it b. Now notice that b(n+1)=a^b(n). Since b(n) goes to b as n goes to infinity, by continuity of the exponential function, get b=a^b.

Note that b=a^b is solvable for b when 0<a<=e^(1/e), and uniquely so when 0<a<=1 or a=e^(1/e) (two solutions otherwise). Graph F(t)= ln(t)/t to see this (asking if F(b)= some constant (ln(a)) is solvable).

[VeryEvilPettingZoo]

So if a^a^a^... exists, then if its value is called b, it follows that b=a^b.

Since b=a^b has solutions for b in terms of a, when a varies, the value of b varies. This creates an implicit functional equation (for x>1, it looks like you have branching choices for y(x). Ignore this complexity). So define y(x)=x^(x^(x^... . For any x in the domain of y (where that weird thing is defined), have y(x)=x^y(x). Now you use can use implicit differentiation.

[VeryEvilPettingZoo]

Wow - check that out. Some programmer at Youtube had a brain fart and started converting characters into html in these comments.

[Jtking3000]

thats a pretty cool problem made rather straight forward. thanks again salman :-)

[YouXKneekXIT]

@ I can help with this...