Derivative of x^(x^x) | Taking derivatives | Differential Calculus | Khan Academy
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- Опубліковано 9 лют 2025
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Differential calculus on Khan Academy: Limit introduction, squeeze theorem, and epsilon-delta definition of limits.
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The base-10 log function is just ln(x) / ln(10). Either base would work for taking the x out of the exponent, but the derivative of ln(x) is 1/x, and the derivative of log(x) is 1/(x * ln(10)), so using the natural log avoids introducing a purposeless factor of 1 / ln(10).
Great video! Especially using the HQ. I enjoy watching these videos a lot.
now second derivative.
no pls ;-;
Dude, you are my hero. I giggled at the rediculousness of it all, but somehow completely understood it. Just lovely.
I can watch these videos in HD all day. Thanks for making new vids!
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
Your writing with the mouse is great . :D
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!
How did you perform 🤔
Thank you so much sir. I have been seeing different videos on this, but i didn't understand anyone of them. But after watching this video, all my doubts regarding this question are cleared. Thank you so much sir.
0:00 - 3:00 - Derivative of y= x^x. Awesome
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.
Great video, very easy to understand.
fantastic problem, ever better solution
@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 ...
thats a pretty cool problem made rather straight forward. thanks again salman :-)
Thanks for the tip of the ln y, greetings from Costa Rica.
really helpful!thanks!
Bars
In my opinion it is much simpler to just write x^x as e^(x * ln x) and apply the chain rule...
thanks salman .
Thanks for posting these videos!
Hey Sal there's actually a rule for solving the derivative of a function to the power of another function (u^v):
d/dx (u^v) = v u^(v-1) du/dx + ln(u) u^v dv/dx
Excellent.
YEAHHH!!! I got it right...
Thanks Sal!
yeah it is because exponents come first in order of operations. He just put the parentheses so it was more clear.
TY, great vid.
Of course, x > 0. If x < 0, then x^x is a multi-valued
complex function; if x = 0, then F(x) = x^x can be made continuous by defining F(0) = 1, but the F'(0) from the
right is -infinity.
You are the best.
rewrote x^x as e^(xlnx) and miraculously got the same answer (charted it in desmos). Good times.
anyone knows which tool he is using to write ? i love the way the letters look , dunno why i just love it and wanna know what he is using
you are my hero .. thanks :)
Thanks!
why take the natural log of both sides? why not just the regular log?
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
woah I get it now thanks!
I wanted to see multiple logging
you are right
you don't understand limits. When taking limits you're not supposed to replace the variable by the value it's tending to. 0^(0^0) is not a defined number, but you can try to see if the limit exists.
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
thanks
There are two ways to solve this thing. First is to take log both side of the function and then differentiate and multiply both side by y like you do.
The second way is my way. First you have y=x^x, this is a function that we cannot deal with because there is no differentiation rule. In fact, there is no formal name to this function. But it looks like exponential function. Yes, it can be changed into an exponential function!
How can we do that? Well, you should know that log and expo are inverse to each other so e^ln (SOMETHING) is equal to SOMETHING
So y=x^x become y= e^ln(x^x) = e^(x ln x)
Now you turn an unnamed function into a familiar exponential function. Just differentiate it and you don't need to move y to any side!
Very nice
Is not better first take the absolute value of both sides and after use natural logarithm? Then you can derive even in negative value of "y". Am i right? Or this crazy expression does not take negative value?
If the movie Inception were a math problem.
brilliant
When he takes the natural log of both sides, isn't he restricting the domain of the function to positive numbers?
thats not necessary if you want the domain also concluded imaginary terms there
x^x implies that x has to be positive. When i plotted it showed me graph on positive x side only.
I can find values of some negative numbers like (-2)^(-2). But not for negative fractions. Im still looking for better explanation.
What is this function called is it an exponential function?
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.
x^5
5x^4
20x^3
60x^2
120x
120
0
...Derp!
why natural log vs using log?
I am in exam. You're crazy man
what is the derivative of 3^(4x)??
the limit is 1
Awesome video! Thank you for posting!
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
salman i graduated from college just cause of you......
de Broglie don’t be slow
nice
@calvinhobbesliker2 Sure, it can. The anti-derivative only can't be expressed in terms of elementary functions.
Lol Decorating is crazy. And this expression. OMG!
But cant i deduct your rule with what he teaches in this video? I think i can. D=
What's the difference between x^(x^x) and (x^x)^x? I know its a bit elementary. but help will be apriciated :)
(x^x)^x = x^(x*x) = x^(x^2)
isnt the dy/dx of x^x just (lnx + 1)
Eddie Swierczewski it's close but there's something missing
I used chain rule x^(x^x) and I got different answer. dy/dx =x^x(lnx+1)(x^(x^x))(lnx^x+1)
d / dx [ x^x ] = x^x ( ln x + 1 )
so how did you apply chain rule and get that answer ?
I love Sal
really helpful! thanks :)
It just doesn't. I tried to use an online integrator and it failed
@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...
New challenge: Integrate x^x (good luck)
Cosmic Paudel does it have an answer? because I'm really curious ~^~
int [ e^( x ln x ) dx ]
There is no answer, it's undefined
(x/(x+1))^x+1
3:30 dont need brackets
He did it to make it more clear dummy
@@allanhenriques2694 7 years late bucko
@ I can help with this...
Find the first derivatives of y=x^(x-2).
can someone shoe the way? i want to compar it with my answer
Yaaaaayyyyy!!!!!!!!!!!!!!!
Ah, I see. His older videos seemed to have bad quality. Which is probably when he was just using mouse.
we can also apply " y=e*(logx*x)= e*(xlogx) and then apply chain rule
X° derivation kaise kare
he probably has a wacom tablet
🥰
That integral can't be done.
don't think that [x^x} is different from x^x
i think there is a mistake at 6:30 X^x(lnX+1) is NOT the derivative of X^x, it is the derivative of Y = X^x... i think the derivative of X^x is just (lnX+1)
raulrobben10 no... You're wrong it is the derivative of x^x
@Dave67004
not that hard, no need to understand a rule from just 2 functions, you differentiate x^(nx) and ur done ^_^
উদ্ভাস জীবনটা খেয়ে দিলো :|
can someone answer this please x^x^x^4
thanks in advance :)
there's nothing to answer
S
No man... we actually want to see it being reduced by naturals logs.... because in an exam they test the method of solving and you won't be able to refer to that (x^x) derivative solution as solved previously.... sigh disappointed...
Huh?
does he write with a mouse???
L
bad does not rhyme with graph
i love you bro (no homo)
why isn't the derivative of x^x just itself? I understand the reasoning here. but it seems like the power rule would yield d/dx(x^x)=x * x^(x-1) which just becomes x^x again. is this just a special case or is there something in the definition of the power rule that doesn't allow it to apply here?
That makes sense, thanks!
thankyou
S
Thanks
S