fake derivative product rule is crazy! 

This formula has a better life than me, at least it has gf.

let g(x) and f(x) be a constant Power Move

If you further require that f(x) = g(x) you get c*exp(2x), which of course works, but is less impressive.

My finals are over, but I keep watching these videos. What has my life come to?

Intersting idea. The (fg)'=f'g' condition is reduced to a easy differential equation.

This sort of thing is used in pretty much every demonstration of ordinary differential equation formulas. It's a good introduction to that.

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

0:00 If?

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

I think I saw this question in Stewarts Calculus textbook

What about the constants of integration?

so glad you got rid of background music in future videos. this was painful

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.

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.

d/dx(1*1)=0=(0*0) It pops out in my brain

Basically, you treated it as a differential equation...

Where is Dr.Peyam at 10:24. USE THE CHEN LU

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.

Introduction to differential equations ? Chain rule is always so nice

Your videos are incredible, I can't stop watching them :D

This sounds like a teacher troll. Part A, derive with product rule. B find f’g’. Part C explain.

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.

If I become a teacher, I will use this to design questions for the tests

Bprp, I support you through this difficult time!

Gauss' Law of Quadratic Reciprocity

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

Why is the constant of integration not necessary?

I think you've not be carefull not dividing by 0 : for exemple with f=e^x

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?

It's like evaluating special integrating factor.

The music is so happy

Asombroso Bellísimo Precioso Maravilloso Inigualable Perfecto

that was so freaking satisfying

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.

how about the fake quotient rule?

5:48 AAAAAH!! My eyes!!!! 😵 You used the same letter for the argument of g and the differential in the integral 😵😵😵

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.

what if f(x) is e^x, the denominator turns 0

f(x)=g(x)=e^(2x) should work

Please do fake matrix product. (When the fake product of two matrices (point wise product) is equal to the actual product)?).

can you try with d/dx(f/g)=f'/g'?

Why shouldn't there be a constant after finding the antiderivative of 1/(2-x)?

Does this formula works for any f(x) that the integral converges???

Why you don't take integration constant

Even the dimensions do not work. First is something /dx. Second is something else/(dx*dx)

When you chose f(x)=x^2: could you just choose any f(x)? Would it be valid for any function?

He missed the arbitrary constant from the indefinite integral. Any function g can be multiplied by an arbitrary constant and it will still work.

Heard that (fg)' = f'g' is called the "freshman's dream", I think on Michael Penn's page... ever collaborate with this maths homie?

What's the name of the background music?? Plsss

If f(x)=exp(x), the rule does not apply because f(x)=f'(x). In this case the denominator vanishes in integral.

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.

Thank you for this video ...

I found this fascinating, could a similar method be used for other problems, say, f'=(f×g)/((f×f)+(g×g))?

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!

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 ^^

f(x)=1 g(x)=1 Done.

how about let both f and g be a constant, then (f*g)' = f' * g' = 0

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?

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

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*/

Any constants would work

Sneaky edit at 6:54, I see you

Super cool

Can't ∫ f '(x) / [f '(x) - f (x)] dx be simplified?

How about the Fake Quotient Rule? (f / g) ' = (f ' / g) + (f / g ')

Insanely cool

f(x) = a, g(x) = b

Thats super cool

How about f and g are both 0?

me inmediately being dumbsmart: f(x)=0, g(x)=0

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.

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

That’s not the only solution. There’s something like f(x)=0, g(x)=0.

Brother you have broken math for all the right reasons haha.

:O

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

But can you do d/dx of f/g is equal to f'/g'?

f and g are constants, everything zero

If both equations are any arbitrary constant, that’s true, correct?

What I can't stop thinking that this is pointless?

Checkout the magic when you use f(x) = e^2x to find out g(x)

Good! Please get rid of that background “music”!

The best solution

Is it actually possible to relate those 2 functions f(x) and g(x)? I wonder...

What if I set f = k f' and g = (1-k) g' ?

That's a very good way for a teacher to evaluate derivatives hahaha; the same answer, but way different process

Craaaaaazy

how about f(x)=g(x)=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

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?

No, you can't pick an arbitrary functions like that. 😭

Magic!

Why didn't you substitute (2-x)^2 with (x-2)^2?

dont forget +C in integral

YAY!

But nobody will know that f(x)=x^2

Great

pura idiozia

I came from the linear differential equation