Shortcut For Derivative of x^x

Here we will take the derivative of x^x FAST
Regular video: • derivative of x^x
------------------------------------------------------------------------------------
👉 Subscribe here: bit.ly/3wvjVL3
🛍 Shop math t-shirt & hoodies: bit.ly/3uZ51vz
💪 Become an undefound member for only $0.99/ month & watch all my unlisted videos first: bit.ly/3uYHrz1
------------------------------------------------------------------------------------
bprp #fast

[@user-yv8mp6kn5f]

the multivariable side of calculus is a pathway to many abilities, some consider to be unnatural

[@gudu12321]

אח יקר

[@sirface7951]

אח בדם

[@ozderei6607]

שלום לכולם

[@cheddlescheeseburger533]

Is this even multi variable calculus?

[@arbs-5164]

​​@@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

[@lasslos1490]

Just do x^x=e^(ln(x)*x) and differenciate as usual. That's, in my opinion, way easier and mathematically complete.

[@ayushraj233]

What will you do in f(x)^g(x) Power mein 😢

[@noobcuber3921]

@@ayushraj233 that isn't f(x)^g(x). it's f(g(x))

[@dhdusidjdjso]

I don’t think this was meant to be a way to solve equations like these, but rather just interesting to see

[@BasitCeviriler]

Yeah, that is what I did.

[@danny89620]

​@@PaulZen30 same goes for the solution in the video then.

[@AyushSingh-pq3cj]

When Eminem becomes a maths teacher:)

[@gaycat599]

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)

[@piyusharora5327]

Thank you for providing this solution

[@StevenMinecraft25]

This is the right way. Logarithmic implicit will get you the most right answers with the least random memorization

[@kemosabe551]

Yeah, in the video he treats the derivative of x^x using something like the product rule, without proper background or justification.

[@noobgaming-eu3or]

Absolutely correct way,here in the equation x^x both the x are variable so one can’t assume them as constant and differentiate

[@Practicalinvestments]

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

[@JamespaulChigozie]

He talks faster than our teacher and yet he makes us understands easily

[@samueldeandrade8535]

You do realise he edits the video to go faster. Right?

[@JGHFunRun]

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

[@user-np4rr1pc3r]

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)

[@gaetanbouthors]

easiest is just x^x=e^(xlnx) so the derivative is (xlnx)'*e^(xlnx)=(lnx+1)x^x

[@thatonesponge4849]

Man i used to love differentiation in Calculus. Integration is what messes everyone up 😂

[@aswinr9676]

How could you cut it off at " and.. we are.. done!!"

[@DouglasZwick]

It really drives home how fast it is when he does that. Do we really _need_ to hear him say the N sound?

[@nanamacapagal8342]

Worse: no marker smack on the whiteboard.

[@PunitKumar-kf3vj]

SIR WHAT IS THE DERIVATIVES OF 1) 5 ^ x+1 2) x^ x+1 3) ( x+1) ^ x❤❤❤plzz batana sir

[@deepapattnaik3454]

Thanks it helped a lot ❤

[@htikeaungwai5513]

Would you accept this challenge? 🤭 Find d/dx (x!) 😈 (Pls make a video of it 😁, wanna know about digamma function)

[@chessematics]

There is already a video on that topic

[@dipsacedemy]

Give him some time , he charming the expansion of x factorial 😂😂

[@random22453]

X! Isn't a continuous function

[@PXO005]

​@@random22453 although the gamma function is, and it is a continuous extension of thr Factorial

[@gonzalesorgaz735]

Use the product rule 🤣🤣🤣

[@ChaineYTXF]

I write it down as e^(xln(x))

[@fredericchalard6611]

In fact, that's exacly how I solved it the first time.

[@athenovae]

Why is the exponent in x^x treated as a number at first?

[@fredericchalard6611]

​@@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.

[@JSFG80]

Love your vids

[@naumovnikolay777]

Thank you!

[@colinjava8447]

I actually remembered all of those derivatives, even the last one.

[@randhyLeksu7288]

the last one got me so 💀💀💀💀💀💀

[@tanushjain7679]

Wow thanks I was it took me a lot of time to do the third one before

[@jtris01]

I alwats did the last one using implicit differentiation

[@theimmux3034]

Is it a coincidence that this works?

[@altarius44]

Nope. That's just how it works lol

[@amethyst2448]

Nothing in mathematics is ever a coincidence

[@theimmux3034]

I've never heard of such a differentiation rule. I usually just convert the base to e and do the usual business.

[@jigobiyoyo4006]

@@amethyst2448 wrong. X^2=25 cancel the 2 twos and you get 5 😁

[@martind2520]

This is the chain rule for partial derivatives.

[@mayureshious]

But how it works. Any logical reason. Plz explain. Does it work for all f^g type

[@pp81191]

I did the exact same thing in school around ~14 years back and showed it to my Maths teacher. He said this is wrong.

[@hotncold7065]

Thanks ☺

[@Hola-rx1jy]

In school we learned logarithmic differentiation which make the second and third easy

[@abdosalah3832]

pretty cool

[@suparnajitghosh7865]

God bless you 😇🙏🏻

[@vnmgaming5172]

Where are our shortcuts

[@mingza8132]

😱🙏 thanks idea

[@leandroteles7857]

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)

[@tunistick8044]

How do I convince yall that it's real

[@nYEOSUh]

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

[@flynner4805]

why do you multiply by ln?

[@Wutheheooooo]

Eminem has been very quite since this dropped

[@ajayvarshney2117]

Is this the superposition of finding derivatives ?

[@not_vinkami]

No, it's just product rule

[@not_vinkami]

@@dp-zn8bd yes it is, kind of

[@not_vinkami]

@@dp-zn8bd x^x=e^(x lnx)

[@yoavshati]

@@not_vinkami But he didn't use this... What he did was find the derivative of f(x,y)=x^y, with x=y

[@thewitchking2556]

@@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)

[@op-gamer3668]

He is only person who speaks like doctor handwriting 😅

[@rahilkeshav313]

Does this work for x^^3 ?

[@makisekuurisuu]

Cool!!

[@littleretroship6403]

That last one was right but it hurt my brain

[@physicsdemonstrations2426]

Wait,!! Is there some other way?!! I would do it the exact same way!

[@secretnobody6460]

What is ln?

[@kazuXZ69]

Bro start rapping

[@RubyPiec]

i don't get d/dx(2^x), isn't that also just 2x? or would that be d/d2?

[@MINEXKILLER]

Done.

[@tzonic8655]

Hahah someone had to finish this

[@MINEXKILLER]

@@tzonic8655 It's an essential part od the video

[@SoulFrmTitanic]

Why am I here ?? Hey bro you should join schools here in India you’ll feel like a genius in integrals

[@mriduljoy8108]

Lool, nope, this is basic CBSE lvl stuff

[@Dark78423]

After watching this i forgot 1+1=2 for some moments

[@markoj3512]

Fun fact this: this video doesn’t show the steps only the solution. The didactic value tends to zero

[@InfinityPie-bw7tc]

3rd one is epic way 😊

[@TheChaos3DS]

Oh he has a beard now

[@1st2nd2]

Minimum of x^x?

[@all_bright]

Taking log on both sides is much easier. 😂

[@Luffy_wastaken]

d/dx ²x = x^x(1+lnx)

[@1badjesus]

HANG ON!... I'm still drawin CIRCLE around #3..

[@_Artistic_journey-]

Am I watching the short in 2x speed??

[@AVIGOOO]

For the third one couldn't you just write it as =>x ln x?

[@justanotherepicrider3974]

X to the X pOwEr 🐓

[@johnnolen8338]

bprp, if times ever get tough, you definitely have a second career ahead of you as an auctioneer, my friend. 😂

[@sushilkumarlohani6709]

I like how he is speaking without breathing

[@mattewdb2321]

And thats how ez math is

[@Optimusprime-ve7xi]

One simple question. What is this?

[@AVIGOOO]

Differentiation, calculus

[@Optimusprime-ve7xi]

@@AVIGOOO ok

[@AVIGOOO]

@@Optimusprime-ve7xi yea prepared to go through hell if you decide to take maths in 11th and 12th Atleast in India

[@ace5161]

​@@AVIGOOOI am in 10th and good at maths so.....

[@replicaacliper]

this is just like multivariable chain rule in disguise i guess?

[@pavelmadarcik3240]

@@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?

[@volxxe]

@@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!

[@helatard4272]

0:00

[@rojansuwal6431]

What abt X^1/x

[@undefeatableyt7]

This easy method is taught in class 11 in India

[@AbhilashSatpathy]

In indian coaching it is in class 9th lol

[@hritakshtripathi3701]

So what?

[@rafiihsanalfathin9479]

And then?

[@undefeatableyt7]

@@rafiihsanalfathin9479 Abe mere najaayaz bete jaake padhai kar varna gäñd par 100 kode bajaunga aur school ki fees bharna bhi band kar dunga teri

[@AstroVp]

yo the third one is different method i solve theses type of qns by taking log in both sides

[@codeovercode167]

Oh yes.. the superposition principle

[@zhangyuchen2000]

partial derivative

[@channelbuattv]

or x^x(ln(ex)).

[@TheStarDreamer]

Does this work for any type of function... Or just x^x ?

[@Elcompalui712]

Any

[@NitrogenTheBlackCat]

How did that work? I don't get it

[@arbs-5164]

​@@NitrogenTheBlackCat 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

[@thatomofolo452]

W 👍👍

[@jay_stack1270]

You’re toooooooo much Bro..... 😳

[@oscarmartinpico5369]

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)?

[@Why_Broken]

Who else is focusing on his beard 😂😂

[@mauriciomartinsdesousa]

Não entendi o que ele disse, mas entendi o calculo.

[@notboboi9977]

bro had to talk so fast💀💀

[@chessnoob9150]

What is the derivative of |x| ?

[@That_One_Guy...]

Clue : |x| = sqrt(x^2)

[@internetexplorerchan2697]

@@That_One_Guy... This property always messed my mind... Makes me confused on (sqrt(x))^2 and sqrt(x^2)

[@internetexplorerchan2697]

@@That_One_Guy... oh yeah it make sense because sqrt(x^2) is x E R so |x| while (sqrt(x))^2 is xER | [0, + inf.].

[@zakiabg845]

The last one is faulse.

[@omarsameer4735]

calculus II

[@juanixzx]

Eso en mi pueblo le llamamos volear machete ventiao

[@adrianhdz138]

Hola yo también hablo español

[@rarebucko]

Isnt that how you're supposed to do it

[@user-hn3wp3sd1m]

Can u do final exam for me I can pay u

[@ammarbasrah1362]

Prove the third rule, please

[@ssdd1316]

and then we are...

[@megalomaniac6464]

You forgot to add c

[@mryip06]

Are u serious?

[@megalomaniac6464]

Yes 😐

[@mryip06]

@@megalomaniac6464 he is not doing integration, no C pls

[@blch5351]

Did he just using 2 pen with one hand LMAO

[@carultch]

Yes. That's the namesake of his channel.

[@teampigs1188]

how is this legal

[@user-wf9xz3tm2b]

Watching in 2x

[@banglaextremefunny2593]

2x speed in this video

[@Jack_Likesmath]

Other people : you can also use this … or this … or easiest like this … Me : …………………… What the fu-

[@Axurb187]

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)]

[@geetharani3122]

Is it just me or is he rapping

[@manas.mk47]

Hindi ke liye 2 dabay!😢

[@arshadalijanwariarshadalij5319]

How are you sir. Prove that , Why derivative of constant is equal to zero ...

[@carultch]

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.

[@mithunghosh3546]

Cant we just cancel the x

[@carultch]

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".

[@archangel1309]

WUT 😃

[@magidgaming5775]

Sir slow down I don’t get what you’re saying

[@lawangenkhan7477]

I don't get it plz..

[@jotadiolynedicci3361]

If you don't understand, just use the fact that a number a^n can be written as exp(n×ln(a)). So doing the derivate x^x is the same as doing the derivate of exp(x×ln(x)).

[@lawangenkhan7477]

@@jotadiolynedicci3361 ok now i get it

[@herbcruz4697]

@@jotadiolynedicci3361 OR start off by writing the function as y=x^x, and then use Logarithmic Differentiation.

[@jotadiolynedicci3361]

@@herbcruz4697 yeah you can do that too.

[@user-ox6vd3lf9q]

Madar aur confused hi gyi sala kl exam h 😢

[@Demonysticx83]

SPEAK SLOWER FFS WE ARE NOT ALL SPEAKING FLUENT ENGLISH MY GOSH