Shortcut For Derivative of x^x
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- Опубликовано: 2 окт 2024
- Here we will take the derivative of x^x FAST
Regular video: • derivative of x^x
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bprp #fast
the multivariable side of calculus is a pathway to many abilities, some consider to be unnatural
אח יקר
אח בדם
שלום לכולם
Is this even multi variable calculus?
@@cheddlescheeseburger533 he basically turned the problem into a multivariate problem and used a calc 3 derivative technique then set the 2nd variable to be x. And by that u basically get the same result as doing it by normal methods.
F = x^y
Taking the derivative wrt x
F'= Fsubx.dx/dx +Fsuby.dy/dx
(Fsubx and Fsub y are partial derivatives of F with respect to x and y respectively)
F'= y.x^(y-1) + (x^y).ln(x)dy/dx
If y=x
F' = x.x^(x-1) + (x^x).ln(x).dx/dx
F' = x^x + (x^x).ln(x)
Which is the form given in the video
Just do x^x=e^(ln(x)*x) and differenciate as usual. That's, in my opinion, way easier and mathematically complete.
What will you do in f(x)^g(x)
Power mein 😢
@@ayushraj233 that isn't f(x)^g(x). it's f(g(x))
I don’t think this was meant to be a way to solve equations like these, but rather just interesting to see
Yeah, that is what I did.
@@PaulZen30 same goes for the solution in the video then.
When Eminem becomes a maths teacher:)
I just assumed it was y, took ln on both side, and differentiated both side with respect to x
y=x^x
ln(y)=x ln(x)
dy/dx * 1/y = x/x + ln x
dy/dx = x^x (1+ln x)
Thank you for providing this solution
This is the right way. Logarithmic implicit will get you the most right answers with the least random memorization
Yeah, in the video he treats the derivative of x^x using something like the product rule, without proper background or justification.
Absolutely correct way,here in the equation x^x both the x are variable so one can’t assume them as constant and differentiate
Oh my lord thank you I did exactly what you did a while ago and wrote the formula in my notes but not the solution so I was mad confused trying to figure out how the hell I got that
Logarithmic differentiation also works.
Y = X^x
Ln Y = ln X^x = x(ln x) *exponent rule with logs
Y'/Y = ln x + x(1/x) *product rule
Y' = Y(lnx + 1)
Y' = X^x(ln x +1)
This way is definitely a simpler way of doing the problem for me
Took me a moment to realize (ln Y)' = Y'/Y due to the chain rule. Well done!
He talks faster than our teacher and yet he makes us understands easily
You do realise he edits the video to go faster. Right?
Would you accept this challenge? 🤭
Find d/dx (x!) 😈
(Pls make a video of it 😁, wanna know about digamma function)
There is already a video on that topic
Give him some time , he charming the expansion of x factorial 😂😂
X! Isn't a continuous function
@@random22453 although the gamma function is, and it is a continuous extension of thr Factorial
Use the product rule 🤣🤣🤣
I was like “Wait that’s legal?!” and then I realized it’s equivalent to differentiating x^y against x and y and then setting x=y
SIR WHAT IS THE DERIVATIVES OF
1) 5 ^ x+1
2) x^ x+1
3) ( x+1) ^ x❤❤❤plzz batana sir
easiest is just x^x=e^(xlnx) so the derivative is (xlnx)'*e^(xlnx)=(lnx+1)x^x
How could you cut it off at " and.. we are.. done!!"
It really drives home how fast it is when he does that. Do we really _need_ to hear him say the N sound?
Worse: no marker smack on the whiteboard.
Thanks it helped a lot ❤
In school we learned logarithmic differentiation which make the second and third easy
I prefer to use this more general shortcut, you could even call it a rule:
let f and g be functions:
dx(f^g) = f^g*(g'*ln(f)+g*f'/f)
In this video's case:
dx(x^x) = x^x*(1*ln(x)+x*1/x) = x^x*(ln(x)+1)
The proof is done by applying logarithmic derivative to f^g
Man i used to love differentiation in Calculus. Integration is what messes everyone up 😂
How about the integral of that
In fact, that's exacly how I solved it the first time.
Why is the exponent in x^x treated as a number at first?
@@athenovaeI fact, here we are evaluating the multivariable derivative of u^v and then we take u=v=x.
Multivariable derivative works like this (D is the multivar derivative)
D(u^v)=d(u^v)/du+d(u^v)/dv
First term we consider v as a number, second term we consider u as a number.
In the short, bprp does that but with the single variable x. It's technically legal.
Respect to bprp for numbing it down for me
Unlike many "tricks" that this channel shows that really are coincidences, this one is not a coincidence, it's actually a valid way to differentiate x^x through the use of multivariable calculus (pretend it's x^y and take the total derivative)
Why am I here ??
Hey bro you should join schools here in India you’ll feel like a genius in integrals
Lool, nope, this is basic CBSE lvl stuff
how bout if it is raised to 2x
d/dx ²x = x^x(1+lnx)
God bless you 😇🙏🏻
How do I convince yall that it's real
why do you multiply by ln?
Is it a coincidence that this works?
Nope. That's just how it works lol
Nothing in mathematics is ever a coincidence
I've never heard of such a differentiation rule. I usually just convert the base to e and do the usual business.
@@amethyst2448 wrong. X^2=25 cancel the 2 twos and you get 5 😁
This is the chain rule for partial derivatives.
Am I watching the short in 2x speed??
He is only person who speaks like doctor handwriting 😅
Fun fact this: this video doesn’t show the steps only the solution. The didactic value tends to zero
HANG ON!... I'm still drawin CIRCLE around #3..
I write it down as e^(xln(x))
For the third one couldn't you just write it as =>x ln x?
I alwats did the last one using implicit differentiation
Done.
Hahah someone had to finish this
@@tzonic8655
It's an essential part od the video
Bro, a very quick friendly talk. Shave that beard , just do it and dont think about it. do it. you will not regret . Trust me
Bro start rapping
Eminem has been very quite since this dropped
We can take x^(x)=e^(xlnx) and just get it easily by chain rule
After watching this i forgot 1+1=2 for some moments
😱🙏 thanks idea
One simple question. What is this?
Differentiation, calculus
@@AVIGOOO ok
@@Optimusprime-ve7xi yea prepared to go through hell if you decide to take maths in 11th and 12th
Atleast in India
@@AVIGOOOI am in 10th and good at maths so.....
Other people : you can also use this … or this … or easiest like this …
Me : …………………… What the fu-
And thats how ez math is
yh i think i will jst become a politician.... this math is crazy
I wish I knew that shortcut for differentiating X^X
Is this the superposition of finding derivatives ?
No, it's just product rule
@@dp-zn8bd yes it is, kind of
@@dp-zn8bd x^x=e^(x lnx)
@@not_vinkami But he didn't use this...
What he did was find the derivative of f(x,y)=x^y, with x=y
@@yoavshati just curious not in that math yet but would a viable way of getting f(x, y)=x^y be (e^lnx)^y =e^ylnx so e^ylnx (y/x + y’lnx)
Where are our shortcuts
SPEAK SLOWER FFS WE ARE NOT ALL SPEAKING FLUENT ENGLISH MY GOSH
Please take it easy...i can't watch your videos
Taking log on both sides is much easier. 😂
Sir slow down I don’t get what you’re saying
What is ln?
Wait,!! Is there some other way?!! I would do it the exact same way!
Isnt that how you're supposed to do it
this is just like multivariable chain rule in disguise i guess?
@@volxxe I think you can, don't you?
"BPRP method":
([f(x)]^[g(x)])' = g(x)*[f(x)]^[g(x)-1]*f'(x) + [f(x)]^[g(x)]*ln[f(x)]*g'(x)
y'= [f(x)]^[g(x)] * [g(x)*[1/f(x)]*f'(x)+ g'(x)*ln[f(x)]]
That's the same as "implicit derivation method":
y=[f(x)]^[g(x)] / ln
ln(y) =g(x)*ln[f(x)] / '
(1/y)*y' = g'(x)*ln[f(x)]+g(x)*[1/f(x)]*f'(x)
y' = [f(x)]^[g(x)] * [g'(x)*ln[f(x)]+g(x)*[1/f(x)]*f'(x)]
And that's the same as the previous.
Am I correct?
@@pavelmadarcik3240 looks like you can, I said you couldn’t because I remembered me trying this on a different function than x^x maybe I just derived it incorrectly anyway thanks!
This easy method is taught in class 11 in India
In indian coaching it is in class 9th lol
So what?
And then?
@@rafiihsanalfathin9479 Abe mere najaayaz bete jaake padhai kar varna gäñd par 100 kode bajaunga aur school ki fees bharna bhi band kar dunga teri
I didn't get ..
But it sounds good😂😂
how is this legal
This may sound complicated but it works
[ f(x) ^ g(x) ] ` =
f(x)^g(x) [ g(x)•ln(f(x) ] `
=
f(x)^g(x) [ g(x)•f `(x)/f(x)+ln(f(x))•g`(x)]
3rd one is epic way 😊
bro had to talk so fast💀💀
Do you not think it might be better if you do y = X^x => ln(y) = X·ln(X) => y'/y=ln(X)+1 => y' = y·(ln(X)+1) => d/dx (X^x) = X^x·(ln(X)+1)?
Easy question 😅😅😅😅
This is cheating
i don't get d/dx(2^x), isn't that also just 2x? or would that be d/d2?
Does this work for x^^3 ?
I thought that this was a troll
But how it works. Any logical reason. Plz explain. Does it work for all f^g type
Bro rapping in calculus
or x^x(ln(ex)).
Would that be calc 2 ?
Brilliant! Thank you for breaking it down like this - making it more digestible!
bprp, if times ever get tough, you definitely have a second career ahead of you as an auctioneer, my friend. 😂
The last one is faulse.
x^x = e^(xlnx)
d(x^x)/dx = e^(xlnx) • (lnx + 1)
= x^x • (lnx + 1)
X to the X pOwEr 🐓
2x speed in this video
I get it now
Watching in 2x
yo the third one is different method i solve theses type of qns by taking log in both sides
Too fast!
Madar aur confused hi gyi sala kl exam h 😢
the last one got me so 💀💀💀💀💀💀
f(x)=x^x, lnf(x)=xlnx, f'(x)/f(x)=lnx+1, f'(x)=f(x)(lnx+1)=(x^x)(1+lnx)
전 로그함수로 바꿔서 푸는 방법만 생각했는데, 저런 방법이 있을줄은 몰랐네요 감사합니다
Who else is focusing on his beard 😂😂
I did the exact same thing in school around ~14 years back and showed it to my Maths teacher. He said this is wrong.
pretty cool
Wow thanks I was it took me a lot of time to do the third one before
Di Matematika Matematika
Oh he has a beard now
Prove the third rule, please
Love your vids
Hindi ke liye 2 dabay!😢
You’re toooooooo much Bro..... 😳
Cant we just cancel the x
No, because the dx in the "denominator" doesn't have anything to do with dividing by x.
"d/dx" is a notation for a derivative, with respect to x. There's some relationship to fractions through the first principles that define the derivative, but d/dx isn't a fraction of d/(d*x), where d is a variable.
d/dx is an operator, not a fraction. and the d's are also operators, and not pronumerals. The d means "infinitesimal change in the following".
I actually remembered all of those derivatives, even the last one.
That last one was right but it hurt my brain
Minimum of x^x?
0:00
How are you sir.
Prove that , Why derivative of constant is equal to zero ...
limit as h -> 0 of (f(x + h) - f(x))/h = f'(x)
Given: f(x) = C
f(x + h) = C
f(x) = C
f(x + h) - f(x) = C - C = 0
Thus, the ratio is 0/h
Because the numerator went to zero before we even started sending the denominator to zero, the derivative of a constant equals zero.