In this video, I showed the relevance and behavior of the Lambert W function and how to compute its derivative
Пікірлер: 88
@ambikachhikara21547 ай бұрын
Hi Mr. Ok! I had you as my Algebra 1 teacher back in middle school and remembered you had a KZbin channel, and now I am in AP Calculus BC and your videos come in handy. It’s great to see that your channel has grown so much!
@blackovich7 ай бұрын
I remember you, Ambika! Good to hear from you! He also taught me Coding. Amazing teacher!
@PrimeNewtons7 ай бұрын
Ambika, that is good to know. Please reach out if you need help. I am proud of your commitment to learning. Never stop learning!!!!!!
@PrimeNewtons7 ай бұрын
You too?!! I am blessed.
@DragonX9996 ай бұрын
@@PrimeNewtonsyou are a goat teacher man
@weo94737 ай бұрын
Next - integration of Lambert w function
@indescribablecardinal65717 ай бұрын
There is a cool equation of an integral of any function given by the integral of its inverse. And the integral of xe^x is trivial 🎉
@rolling_metalmatica7 ай бұрын
Taylor Series Expansion for the Lambert W Function would be cool
@T1Pack7 ай бұрын
0⅘
@Anmol_Sinha7 ай бұрын
@@indescribablecardinal6571do you mean that integral of f(x) wrt x = integral of f-1(x) wrt y? The comment asked for the integral of f-1(x) wrt x. To find the integral we can take the last step in prime newton's video, cross multiply for W(x) and integrate. We will get the answer already mentioned in this comment chain
@user-ct1iv9dq1b7 ай бұрын
There is a formula for integrating an inverse of a function,and W is just an inverse of xe^x,that wouldn't be that hard.
@octs6096 ай бұрын
I do not know anything of calculus, and man I hated math, but for some odd reason, I can not help, but be so intrigued. I blame my educators for me being so bad at math, but also so uninspired and uninterested, after all I was a child, but I commend you for revitalizing my love for math. Your a godsend mate.
@Misteribel7 ай бұрын
The trick you apply by taking the derivative on both sides (9:10), then using the product rule, and get back a component that's itself containing the derivative (W'(x)) really caught me off guard. So simple and so useful! It allows you to find the derivative of the productlog function by inference, using basic high school differentiation rules and never really differentiating the function itself directly.
@PrimeNewtons7 ай бұрын
Great tip!
@looney10235 ай бұрын
Implicit differentiation is really powerful. You can use it to find the derivative of the inverse of any function working solely with the function itself.
@rivalhunters46667 ай бұрын
aah, u forgot the bracket at the end MY OCD IS TRIGGERED. A very good video :)
@deathracoffee7 ай бұрын
I just wanted to say, I really like your voice. Keep on being awesome
@johnsellers58187 ай бұрын
I've taken many math courses up through graduate school and you are the best teacher I've encountered.
@laman89147 ай бұрын
We love how this dude is lecturing Math. Step-by-step. I have watched a number of Lambert W-function clips and they all start right away. But here, you are introduced to the fundamentals first and then how they apply to the actual problem. So, even if you have never heard of it, you can still follow the explanation. We wonder if he has this all hidden in his hat.
@remopellegrino89617 ай бұрын
KZbin needs more Math people like you and Michael Penn
@kusuosaiki3677 ай бұрын
I have watched few of your videos. As a Math student, I really find these interesting. Keep it up good sir.
@johannaselbrun6 ай бұрын
Gracias por apoyarme y me gusta tu trabajo mucho
@Ron_DeForest7 ай бұрын
I have to say that’s an amazingly fast turnaround. Request a video one day, get it the next. Wasn’t quite what I was hoping though. Was really hoping for a deep dive into how it actually works. There’s more to it besides being very convenient. If you use the function on a calculator it comes up with an answer.
@lambertWfunction_Ай бұрын
goated teacher man, great explanation
@biswambarpanda44687 ай бұрын
You are superb sir
@EvilSandwich7 ай бұрын
Thank you. So many people covered this before but they tend to just glaze over a lot of the simplification. Which usually would be fine, but for a function like this, it just feels like their skipping steps and I'm grateful you took your time and explained every step. Any plans to explain how to integrate W(x) in a future video too?
@PrimeNewtons7 ай бұрын
Yes
@koenth23596 ай бұрын
Your teaching skills are beyond normal!
@PrimeNewtons6 ай бұрын
Glad you think so!
@donsena20132 ай бұрын
Quite an analysis !
@ikhsanmnoor85897 ай бұрын
Then I meet this really good explanation
@user-yd4ky5vb3w7 ай бұрын
Thanks for an other video...master
@jadenredd7 ай бұрын
good video today unc 👍🏾
@inceden_Matematik7 ай бұрын
Soo good :)))
@VincentGPT-lol7 ай бұрын
Interesting lesson today 🤓✍️
@ferretcatcher23772 ай бұрын
This is elegant mathematics. ❤ the use of the chalkboard. Reminds me of my salad days at university.
@user-xw6ky8ob4l7 ай бұрын
Admire your love for Mathematics. This runs through your veins. This in turn is a reflection of your love for every learner under your wings. Here we could revisit Kuert Goedel to probe his incompleteness theorem which classifies three possibilities for solutions given Lambert W Function. No solution exists, and new tools are to be discovered. Lambert W Function only offers an endless loop of no empirical value. Stay Blessed.
@richardbraakman74697 ай бұрын
You could also instead of factoring out the e^W(x), replace the W(x)e^W(x) with just x. Then you get 1 / (e^W(x) + x)
@TheLukeLsd7 ай бұрын
eu faço deste jeito também. é mais fácil.
@shshshshsh76127 ай бұрын
for the third version, we see W'(x)(e^W(x) + W(x)e^W(x)) = 1 but W(x)e^W(x) = x by definition, so W'(x)(e^W(x) + x) = 1. so W'(x) = 1/(e^W(x) + x)
@CalculusIsFun17 ай бұрын
Alternatively you could have used the formula for inverse functions derivative based on the regular function. If y = f^-1(x) then f(y) = x 1 = f’(y) * dy/dx Dy/dx = 1/f’(y) y = f^-1(x) Therefore the derivative of any inverse function can be represented using its none inverse counterpart as dy/dx = 1/f’(f^-1(x)) Let apply this to lambert. The derivative of xe^x = e^x(1 + x) so d/dx(w(x)) = 1/f’(w(x)) where f’ is e^x(1 + x) So derivative of the lambert function is 1/(e^w(x) * (1 + w(x))
@senkum10007 ай бұрын
I ALSO MADE THAT FORMULA
@Ferraco057 ай бұрын
The "third" version really just gives you back the first version. On another note, you could write a "fourth" version: d/dx [ln(W(x))] = 1/[x(1+W(x))]
@user-yd4ky5vb3w7 ай бұрын
از شما وبزنا شما متشکرم
@brian554xx7 ай бұрын
) I felt compelled to indicate that.
@priyansharma15127 ай бұрын
Great vid as always but that bracket missing from the second solution has me so annoyed 😭😭
@davefried7 ай бұрын
how would you write the answer in terms of the original equation that the lambert function is based upon?
@giorgiobarchiesi50035 ай бұрын
Tank you for the video! But I wonder if it would make sense using the rule of the derivative of the inverse of a function. If I remember correctly, it should be the reciprocal of the derivative of the function. For a monotone function like this, it should work just fine.
@PrimeNewtons5 ай бұрын
Yes. That works, too.
@jonathanv.hoffmann30897 ай бұрын
🎉🎉🎉
@overlordprincekhan7 ай бұрын
TBH, Another elegant solution would be to use taylor series of e^x and multiplying it with x would give you lambert w function. Then differentiating the series should yield the derivative of Lambert W function
@NekoChan_TV5 ай бұрын
derivative of W(x) is aesy, it's W'(x) ! Apart of that little joke, thanks for sharing us your knowledge !
@mazabayidolazi7 ай бұрын
Good
@chengkaigoh51017 ай бұрын
Is this possible by first principle?
@nanamacapagal83427 ай бұрын
You can use this definition: lim_a->x (W(a) - W(x))/(a-x) Then substitute a = be^b x = ye^y On one specific branch at a time this substitution is okay Then it's lim_b->y (b - y)/(be^b - ye^y) = 1 / lim_b->y (be^b - ye^y)/(b - y) = 1/ (d/dy (ye^y)) So if you can get the derivative of xe^x by first principles then you're all clear This actually generalizes: d/dx f¯¹(x) = 1/f'(f¯1(x))
@wafflesauceyАй бұрын
@@nanamacapagal8342would using this formula cover both of the real branches of the W function?
@anglaismoyen7 ай бұрын
You forgot to close the bracket at the end. Faith in this channel destroyed. Nah, just kidding. Beautiful derivative.
@PrimeNewtons7 ай бұрын
Thanks for keeping the faith 🤠
@aguyontheinternet84367 ай бұрын
12:47 if you did that and cancelled out the W(x) on the top and bottom, you'd end up with the first equation.
@suyunbek13997 ай бұрын
how do you use the derivative of the inverse function formula here? derivative of x*e^x is (x+1)*e^x then what?
@anotherelvis7 ай бұрын
If f(x) is the inverse of W(x), then the formula for the derivative of the inverse gives us W'(x)=1/f'(W(x)) Now insert f'(x) = (1+x)*e^x to get W'(x)=1/((1+W(x))*e^W(x))
@empathy8007 ай бұрын
Instead of writing the solution in terms of Lambert function, could you simply calculate the inverse of the function that is the Lambert part?
@vnms-6 ай бұрын
I just did: W(x) = y -> x = ye^y then derived, so: 1 = dy/dx • e^y + ye^y •dy/dx -> 1 = dy/dx(e^y + ye^y -> dy/dx = 1\(e^y(1+y) Since y = W(x) and dy/dx = W’(x) that means: W’(x) = 1/(e^W(x)(1+W(x))
@RileyGallagher-ce4rq4 ай бұрын
You can also do this: (I'm letting y = W(x) for the sake of not writing W(x) 7 times) dy/dx = (dx/dy)⁻¹ = [d(yeʸ)/dy]⁻¹ = 1/eʸ(y+1)
@DroughtBee7 ай бұрын
I really don’t like how you didn’t close your parentheses at the end on the denominator. Otherwise great video!
@PrimeNewtons7 ай бұрын
🤣 Apologies
@navyntune815813 сағат бұрын
Third derivative: W'(x) = 1/(e^W(x) + 1)
@dhiaguerfi26027 ай бұрын
6:44 f must be bijective
@donwald34367 ай бұрын
Are you related to Omar Epps you could be his brother lol.
@usernameisamyth7 ай бұрын
@v8torque9327 ай бұрын
I don’t watch it for the math. I watch to see a black dude smile and pause it it brings me joy
@salvatorecharney81806 ай бұрын
Because [W(x)]e^[W(x)] is just x, can you write the final answer: 1/(e^[W(x)] + [W(x)]e^[W(x)]) As this: 1/(e^[W(x)] + x)
@lazaredurand6675Ай бұрын
"Never stop learning..." is actualy a wrong slogan because IA can actualy learn non-stop and they will never be living being. The good one would be "Never stop to search/try/be curious". IA will never be curious, curiosity is the proof that you are living.