fake derivative product rule is crazy!
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- Опубліковано 26 вер 2024
- A FAKE PRODUCT RULE? Math for fun!
Is it possible for us to find non-constants f and g so that the derivative of f*g is just f'*g'? In fact, we will end up solving a differential equation to find the answer-generating equation! Enjoy! See here for more math for fun video: • Playlist
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This formula has a better life than me, at least it has gf.
but it always ends up becoming an x
The x will definitely miss the d at last.
Best nerd jokes ever 😂😂
You r great🙏🙏🙏
gf's just become future x's leaving you to ask y you even got with them.
let g(x) and f(x) be a constant
Power Move
daym bro
My finals are over, but I keep watching these videos. What has my life come to?
enlightenment
CAN'T DENT THE KENT just like me. Semester is over, but I am still teaching! :)
CAN'T DENT THE KENT I started doing for fun..
Your life has come to your finals being over and you are watching videos
A good life
Intersting idea. The (fg)'=f'g' condition is reduced to a easy differential equation.
If you further require that f(x) = g(x) you get c*exp(2x), which of course works, but is less impressive.
f = 0 and g = 0 is working.
But during your calculus your divided by some function, so you won't find those solutions !! You won't find any solution that may pass through the absciss axe too :P
c1 and c2 are any real number
f=c1 ,g=c2 works too
@@nepraos3151 no cuz the derivative of those constants are all zero and they are not equal to each other
@@gaveferia1421 think again
@@nepraos3151 *facepalm*
Read it as fg=f'g'
@@gaveferia1421 (c1*c2)'=(c1)'*(c2)'
This sort of thing is used in pretty much every demonstration of ordinary differential equation formulas. It's a good introduction to that.
0:00 If?
Fematika ah, typo!
I think I saw this question in Stewarts Calculus textbook
What about the constants of integration?
They were stolen by the state
It's the price we pay for living in a civilized society.
The constant of integration becomes a constant multiplier for g
One very easy solution is f(x) =e^(ax) & g(x)=e^((1-a)x)
This way f(x) *g(x) =e^x and the derivative is the same
For example f(x) =g(x) =e^(0.5x)
Exponent rule says f(x) *g(x) =e^(0.5x+0.5x)=e^x
d/dx(e^x) =e^x=e^(0.5x)*e^(0.5x)
This would mean you could even use sin²(x) + cos²(x) =1
Where is Dr.Peyam at 10:24.
USE THE CHEN LU
Basically, you treated it as a differential equation...
so glad you got rid of background music in future videos. this was painful
Great!
I tried to find a solution and got some with exponentials, but I was surprised to see that any other function could get a counterpart to solve this equation.
When you derive a polynomial, you decrease of 1 degree in x, here you say that the derivative of the product is equal to the product of 2 derivatives, something that should have 2 degrees less in x. Somehow we know that the global form of f*g has to be of order 0 in x, or be described by an infinite series (infinite order), and that's indeed what we get.
d/dx(1*1)=0=(0*0)
It pops out in my brain
Introduction to differential equations ?
Chain rule is always so nice
If I become a teacher, I will use this to design questions for the tests
Your videos are incredible, I can't stop watching them :D
Bprp, I support you through this difficult time!
?
4:58 Another way to show this works is 1/g (g') dx = 1/g (dg/dx) dx = 1/g dg = d(lng). Take the integral on both sides you get I 1/g (g') dx = I d(lng) = lng.
Gauss' Law of Quadratic Reciprocity
It's like evaluating special integrating factor.
I saw ∫ f/(f-f') dx and thought: oh, cool, it should be an easy integration. I do not have absolute any intuition for integrals hahahaha...
Also, I love this channel. I was never that good at Calculus but here it is really fun to watch.
I really like these math for fun videos; keep it up! Also, I am starting calculus 3 in january. Do you have any plans to go beyond chapter 11 in the Stewart Calculus book?
nothing sorry. But I am teaching DE in spring and will have vids for those
blackpenredpen Sweeeeet! I'll be looking for those next fall!
blackpenredpen I also have differential equations next fall so they will be very helpful to me as well
This sounds like a teacher troll. Part A, derive with product rule. B find f’g’. Part C explain.
The music is so happy
that was so freaking satisfying
I found the following solution
f = Ce^(x/k)
g = De^(x/1-k)
for some constant C, D and k
k(1-k)≠0
I think you've not be carefull not dividing by 0 : for exemple with f=e^x
It's been a few years but this video is really beautiful, I wonder if we can apply this to other cool rules
Duvuvier could have just said that the formula works when either function is 0.
Anyway, what I wanted to say was something different.
Multiplication is a commutative operation (i. e. an operation where the order of the elements doesn't affect the result), so the formula should also work if I exchange f and g.
What I get is:
f=e^int(g'/(g'-g)) dx
Now I will take the ln of both sides:
int(g'/(g'-g)) dx=ln f
Now I can differentiate both sides:
g'/(g'-g)=f'/f
Now I can cross multiply:
fg'=f'(g'-g)
In the right-hand side I get:
fg'=f'g'-f'g
And if I take f'g to the left-hand side, I get:
fg'+f'g=f'g'
Which is exactly the condition you wanted.
Why is the constant of integration not necessary?
Ok, I just answered to myself x'D
@@rodrigoibarraran2133 why tho
@@alejrandom6592
Because everybody knows it's there 🤫
5:48 AAAAAH!! My eyes!!!! 😵
You used the same letter for the argument of g and the differential in the integral 😵😵😵
how about the fake quotient rule?
Just another differential equation...
This only gives solutions for which g(x)>0 and f(x)≠f'(x) for all values of x. But there are many other solutions. For instance, there is the family of solutions (f(x),g(x)) = (k*exp(x),0), where k is any real number. Also, by symmetry, whenever (f,g) is a solution, so is (g,f), even if f is strictly negative. But what are really tricky are the bewilderingly numerous solutions (indeed, almost all of the solutions) for which both f and g both cross the x-axis and therefore cannot be found using the formula here.
Asombroso
Bellísimo
Precioso
Maravilloso
Inigualable
Perfecto
Horrible.
@@luismariabiaggioni8514 What an idiot.
what if f(x) is e^x, the denominator turns 0
zeyuan cao of course f(x)=e^x is not valid. Moreover, there are lots of other functions that cannot suit the equation. This only applies for certain function, such as rational functions, etc.
Please do fake matrix product. (When the fake product of two matrices (point wise product) is equal to the actual product)?).
f(x)=g(x)=e^(2x) should work
can you try with d/dx(f/g)=f'/g'?
A trivial solution (that will often come up if you don’t have constants or some kind of mess) is either is 0 and the other is anything, or they are both constants.
Why shouldn't there be a constant after finding the antiderivative of 1/(2-x)?
Even the dimensions do not work.
First is something /dx.
Second is something else/(dx*dx)
Heard that (fg)' = f'g' is called the "freshman's dream", I think on Michael Penn's page... ever collaborate with this maths homie?
He missed the arbitrary constant from the indefinite integral. Any function g can be multiplied by an arbitrary constant and it will still work.
When you chose f(x)=x^2: could you just choose any f(x)? Would it be valid for any function?
I think yes, as long as it is a function...
It would be valid for any function, that's what he "forced" at the beginning of the video. However, the integral in the exponent would very likely become a pain to compute
Giulia Cardozo Yeah it would, but he chose something simple to integrate in order to get a clean g(x)
Ok thanks =)
that's right
Why you don't take integration constant
Does this formula works for any f(x) that the integral converges???
If f(x)=exp(x), the rule does not apply because f(x)=f'(x). In this case the denominator vanishes in integral.
I also found f(x)=g(x)=e^(2x), but f=g=1,0 also work. didn't know f=x^2 had a solution. Cool video!
I found this fascinating, could a similar method be used for other problems, say, f'=(f×g)/((f×f)+(g×g))?
Yea, we can try it.
For arbitrary g
f =c*e^∫[1/(1- g/g')] will be a solution for a given initial value of f c can be determined and the solution will be unique
how you got this
@@alejrandom6592 It's been 3 years dude. I don't remember
@@turbopotato4575 hhahah
hi .. i love your videos so much but can do more than this if you do some exercises in physics .. or talking about physics rools .. and demonstrations .. like relativity and schrondinger .. etc .. thanks ^^
Thank you for this video ...
Any constants would work
Brother you have broken math for all the right reasons haha.
me inmediately being dumbsmart: f(x)=0, g(x)=0
f(x)=1
g(x)=1
Done.
What I can't stop thinking that this is pointless?
Look at your phone, in fact, look at every piece of technology in your house and ask yourself what would've happened if everyone did think as you did.
19midnightsun87 I was afraid of this to happen that my comment is going to be miss understood. I was just saying that if we could say "ah in that situation we can use that way" than this work would point something.(I didn't say that, that's why I comment that)
I hope I could write correct what I'm trying to say.
İnan Uygur There are tons of applications for calculus. The most basic one is to refine and form your understanding further.
İnan Uygur btw I would just like to add that this kinda approach is very similar to find the integrating factor when we are solving linear differential equations. However, as the title suggested, I did this fake product rule for fun. :)
DE videos will be up next month, if not sooner.
Insanely cool
how about let both f and g be a constant, then (f*g)' = f' * g' = 0
what about the fake chain rule/* is it possible to have d/dx(f(g(x)) be f'(g'(x)) where f and g are non constants because *yawns* that would be bloody boring*/
Sneaky edit at 6:54, I see you
You got that five seconds late chief
Checkout the magic when you use f(x) = e^2x to find out g(x)
Can't ∫ f '(x) / [f '(x) - f (x)] dx be simplified?
N0tY0ur4v3r4g3N0th1ng not really...
That’s not the only solution. There’s something like f(x)=0, g(x)=0.
f(x)=0 and g(x)=any function
f and g are constants, everything zero
How about the Fake Quotient Rule?
(f / g) ' = (f ' / g) + (f / g ')
Or (f / g) ' = f ' / g '
We can do the same with quotient rule. However I worked it out and the result was either crazy or not as interesting.
f(x) = a, g(x) = b
The best solution
can someone explain why 4:43 to 4:50 holds? I've never seen an integral like that but apparently it is something you should know by heart.
He is using the chain rule in 'reverse' by knowing the derivative of the logarithm of another function.
Assume you have the functions u(x) and v(x). Chained they could be w(x)=u(v(x)). The derivative is w'(x)=v'(x) * u'(v(x)).
What happens if u(x) = ln(x)?
u'(x) = 1/x
So the derivative of w(x) is
w'(x) = v'(x) / v(x).
In his video v(x) is g(x). Through experience he knew that the integral of g'(x) / g(x) are the chained functions g(x) and ln(x).
Super cool
Thats super cool
What's the name of the background music?? Plsss
That's a very good way for a teacher to evaluate derivatives hahaha; the same answer, but way different process
At some point in the video he starts dividing with g and f' - f. How can he do that if he doesn't know that they are zero?
If they were zero they would be inmediate answers. The rest of the answers have to be of that form.
Francisco Jesús Morilla Ortega I'm not saying that they are 0 for every value of x.. what if they are zero for some specific values of x? Then he can't plug that value in f or g because he would be devising by 0, thus equivalence is lost..
Andrew Sp yeah he lost Many solutions at this Point.
I mean with his e-Formulia he cant become answers who are 0 an Some points, g>0, but this is just wrong.
Andrew Sp For continuously differentiable functions, it is not possible for the functions to be 0 only at isolated points while also satisfying these differential equations. The functions need to be smooth for the analysis to make sense. Otherwise, they are known as pathological functions, and pathological functions cannot be studied, by definition.
Andrew Sp To illustrate my point, if f'(x = c) - f(x = c) = 0 for some c, then f'(c) = f(c). The only continuously differentiable function family which can satisfy this identity is f(x) = Ae^x. You can probably find some other functions, but then they are not continuously differentiable, and so it does not make sense to even consider the equation we began with, so they are irrelevant to the analysis.
Similarly, if g(c) = 0, all this implies is that the solution family will exclude c from the domain.
But can you do d/dx of f/g is equal to f'/g'?
how about f(x)=g(x)=0
SmileyMPV or any other constants
I just watched this video but I had to turn my speakers volume up really loudly coz you're so quiet ... then when I was done watching your video I switched to another tab where I had a music video on youtube loaded ... I pressed the play button and since I forgot to turn the volume back down I blasted my ears off coz it was so loud
If both equations are any arbitrary constant, that’s true, correct?
No, you can't pick an arbitrary functions like that. 😭
Is it actually possible to relate those 2 functions f(x) and g(x)? I wonder...
James Tantono did you watch the video? That's exactly what he did.
Watch the video first...then comment
Can you please explain why in the part when you integrate g'/g you ignored that integrate of g' is g , so that there should be g*ln(g)? I cant get that part, but i assume that Im just dumb and that is pretty obvious :D
Slidemat because if you do d/dx( ln(g) ) then you'd get 1/g*g' chain rule. So when you integrate g'/g you will just get ln(g)
blackpenredpen Oh key i see now, thank you very much :)) your the fastest responding math teacher lol :P
Slidemat you're welcome. Thank you for watching as well.
Good! Please get rid of that background “music”!
How about f and g are both 0?
And what if you had let g(x)=0 and f(x) be anything you want to wouldn´t it also be an answer
That's what I first thought
dont forget +C in integral
lol doesn't matter in this case, he is only looking for one answer. f = g = e^(2x) is also an answer lol
It's better to write antiderivative of (g'/g) as ln |g| instead of ln g. ln g is good when g is positive. So you have to put absolute value of g instead of g so you have to write + or - before your formula. Also when you will be finding antiderivative of f'/(f'-f) remember it can differ with any constant so you can add any C in exp and it will result in multiplying by any positive constant and with this +- thing it means that you can just multiply it by any constant which is obvious if we look at the fake product rule. If the rule works for any functions f and g you can multiply these functions by any constants and rule stile holds. So this formula for g picks 1 function of 1-dimentional space of all functions that would work. Also this formula works on some neighbourhoud around point x0 where f(x0) is not equal f'(x0) (we assume f' is also continous, we also need it on your formula so we know we can integrate f'/(f'-f) in Newton's sense (finding antiderivative). I was wondering what if f(x)=f'(x) on some interval. Then f(x)=c*exp(x) and that means g(x) has to be 0 on this interval if c is different than 0 or g can be any function when c is 0. Now it would be interesting what happens when f(x1)=f'(x1) in some point x1 and f(x2)=/=f'(x2) in some other x2 and also (f is define on open interval containing both x1 and x2). Will g explode at x2? Or maybe will converge to 0 (since g=0 in case f=c*exp(x))? Maybe both situations are possible and it depends on f? Does it depend on f'' (assuming it exists)?
PS. In your case of f(x)=x^2 we see there are 2 points where f=f' and they are 0 and 2. It turned out you can have 1 continous g that works at surrounding x=0 but rule also holds for x=0. But for x=2 it turned out that g blows up. I think there is also possible third case where g becomes 0 and it's a lil bit different than what we have here at x=0. It would be nice to characterize when all cases happen.
PS. Ok it's clear that 0 was the point of discountinuity of f'/(f'-f) that can be removed and that's why rule hold. x=2 was there the function f'/(f'-f) exploded and became + infinity. Thing is it doesn't mean that antiderivative has to be explode (example sqrt doesn't explode at 0 but its derivative does). So question is how much is changed only by the fact that funtion in integral is of a form f'/(f'-f)? Does it mean that when this function explodes, antiderivative also explodes? Can it have 2 signs depending on side of explosion point?
Wrong, he can only get the positive solution with his fomula exactly because of the expeonential. But you can multiply g with any constant and it will still work.
What if I set
f = k f' and g = (1-k) g' ?
:O
= )
Magic!
Craaaaaazy
It's as fake as (a+b)^2 = a^2 + b^2.
Why didn't you substitute (2-x)^2 with (x-2)^2?
Why would you?
It's standard to put the highest degree term first and make the first term positive.
Lucian Willi no it isnt
@@Rain99891h yes it is. Change my mind.
But nobody will know that f(x)=x^2
Vicente Valdés there are other choices. You can choose f(x)=x^3 and you can work out a corresponding g(x) to make it work. There are inf many choices.
YAY!
I came from the linear differential equation
pura idiozia