Hi Mr. Ok! I had you as my Algebra 1 teacher back in middle school and remembered you had a TH-cam 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!
@@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
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.
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.
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.
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.
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.
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?
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.
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))
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.
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
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))
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))
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))
"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.
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Hi Mr. Ok! I had you as my Algebra 1 teacher back in middle school and remembered you had a TH-cam 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!
I remember you, Ambika! Good to hear from you! He also taught me Coding. Amazing teacher!
Ambika, that is good to know. Please reach out if you need help. I am proud of your commitment to learning. Never stop learning!!!!!!
You too?!! I am blessed.
@@PrimeNewtonsyou are a goat teacher man
Next - integration of Lambert w function
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 🎉
Taylor Series Expansion for the Lambert W Function would be cool
I think the indefinite integral is =
x (W(x) + 1/W(x) - 1) + c
0⅘
@@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
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.
aah, u forgot the bracket at the end MY OCD IS TRIGGERED. A very good video :)
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.
Great tip!
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.
I just wanted to say, I really like your voice. Keep on being awesome
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.
I've taken many math courses up through graduate school and you are the best teacher I've encountered.
TH-cam needs more Math people like you and Michael Penn
This is elegant mathematics. ❤ the use of the chalkboard. Reminds me of my salad days at university.
I have watched few of your videos. As a Math student, I really find these interesting. Keep it up good sir.
goated teacher man, great explanation
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.
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)
Gracias por apoyarme y me gusta tu trabajo mucho
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?
Yes
You are superb sir
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.
Your teaching skills are beyond normal!
Glad you think so!
Quite an analysis !
Then I meet this really good explanation
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))
I ALSO MADE THAT FORMULA
Thanks for an other video...master
good video today unc 👍🏾
Interesting lesson today 🤓✍️
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))]
از شما وبزنا شما متشکرم
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)
eu faço deste jeito também. é mais fácil.
Soo good :)))
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.
Yes. That works, too.
how would you write the answer in terms of the original equation that the lambert function is based upon?
)
I felt compelled to indicate that.
derivative of W(x) is aesy, it's W'(x) !
Apart of that little joke, thanks for sharing us your knowledge !
🎉🎉🎉
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
Great vid as always but that bracket missing from the second solution has me so annoyed 😭😭
You forgot to close the bracket at the end. Faith in this channel destroyed. Nah, just kidding. Beautiful derivative.
Thanks for keeping the faith 🤠
Is this possible by first principle?
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))
@@nanamacapagal8342would using this formula cover both of the real branches of the W function?
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.
Good
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))
Instead of writing the solution in terms of Lambert function, could you simply calculate the inverse of the function that is the Lambert part?
how do you use the derivative of the inverse function formula here?
derivative of x*e^x is
(x+1)*e^x
then what?
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))
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)
I don’t watch it for the math. I watch to see a black dude smile and pause it it brings me joy
I really don’t like how you didn’t close your parentheses at the end on the denominator. Otherwise great video!
🤣 Apologies
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)
Are you related to Omar Epps you could be his brother lol.
6:44 f must be bijective
W function , it just gives you back your ex 😂
"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.
Now, where's the L function 🙃