You sound like a software engineer.

that pun is nice

What was the problem?

Pun intended?

Nice story. But did it actually happen?

not really, rules are much easier to remember as a student

Have you read Lockhart's Lament? Touches on this beautifully.

Well put.

So, the process of exploration is something most of the class won't understand. They're explained only later what all of it meant? I can't get behind that idea at all. Sure, we can give examples of applications, but even then, often understanding the application requires knowledge of theory, as well. E.g how to solve very big systems of linear equations using Banach fixpoint theorem. You won't even understand what it means without theory. On a cynical note, I am getting quite tired of this mantra of "best way of learning something without putting in any serious effort". There is none.

@@feitan8033 It's not *purely* about remembering. It's about understanding. And I find the understanding of concepts is lacking.

Adam Romanov Both. I’ve seen him at the rock climbing gym with his kids before. I’ve also heard things about his pizza. I just graduated as a physics BS,l from Randolph College, so I had him for calc 2, differential equations, and linear algebra. He likes to do calisthenics in class and he ripped his jeans one time in an 8AM DE class.

@@GalaxyGal- From my perspective you a very lucky to have such a professor.

@@GalaxyGal- for real? You are lucky to have him as Professor

@@GalaxyGal- Oh my God, I feel so bad! He does sound like a legendary professor though. Wish I had him as my prof.

Aden Power He’s awesome. He has great chemistry with the other math professors from what I can tell. He got me roped into hagoromo too. It’s a shame the school’s social media don’t talk about his channel that much.

You mean h(t) and integral dt right

Thanks for your comment. I had typos and corrected my comment above.

In this case "everything is nice" is okay. All you need is "differentiability" and "not divide by zero". So it´s okay.

Me too

What is that type of derivative called, so I can look it up?

@@iang0th Yes you can.. It is called the multiplicative derivative.. Every bijection or more generally a homeomorphism gives rise to a pair of calculi.. In this case, the bijection is the exponential function and it gives rise to multiplicative(or geometric) derivative and bigeometric derivative and respective integrals as well.. The idea to is to convert the usual four algebraic operations to a subset of the real numbers using a bijection then use the converted operations to construct these calculi and much more .. A wider scope is Differential Geometry .. However, you can look this up from papers and a book called Non-Newtonian Calculi by Michael Grossman and Robert Katz.. Several good papers were also published on this topic by Agamirza Bashirov.

I'm conjecturing that the rule "(f + g)*(x) = f*(x) + g*(x)" doesn't hold in general. Win some lose some I guess. Also (e^x)* = lim [h -> 0] of (e^{x+h}/e^x)^(1/h). This expression equals (e^h)^(1/h) which equals e^{h * (1/h} = e which is independent of x and h. So (e^x)* = e for all x. This seems like the most expected thing. Conjectured corollary: if f*(x) = c for some c then x is an exponential function, hopefully even f(x) = c^x. Proof left as an exercise for you ;-)

g=g' => f'g+fg' = g'(f+f') = g'f' => f+f'=f' => f=0. So g' can only be equal to g if f=0. And f can only be f' when g=0 for symetry reasons. I'm not very good at this so please tell me if I made a mistake.

I'll also have to add that g=g'=0 is also possible (since I divided by g')

it's a very copout term in math. It is most likely referring to a smooth function (infinitely differentiable). In other contexts, it might be further restricted to analytic. A lot of functions that we by default work with happen to be smooth and analytic: polynomials, trig, exponential

I see @Adriano Andrade , the solution for g(x)=e^ax have a-1 at the denominator so a can't be 1

İt's a mathematician's orgasm

Playing with g(x)=ln(x) made my day, next try: g(x)=log_b(x)

I don´t get you. It is obvious, what "nice" means in this context.

Don't you have a patreon or something?

it's not more truthful - they both differentiate to the same thing; for x

I thought this too.

Not necessarily if you allow to apply functions like ln or sin without any parentheses. Which is actually done in the video, notice the sin 𝑥 everywhere. To elaborate, this would mean you interpret something like sin 𝑥² as sin(𝑥²). For squaring the result instead, you’d write (sin 𝑥)² or some people write sin² 𝑥 (but I don’t like this notation). This means that parentheses like in ln(𝑟−𝑥) are still needed, but only since ln 𝑟−𝑥 would mean ln(𝑟)−x, not because ln always needs some parentheses. Then ln(𝑟−𝑥)⁰·⁵ would be similar to the sin 𝑥² above and it would mean ln((𝑟−𝑥)⁰·⁵). But don’t forget, notation is not universal, everyone can do whatever they like. For example you can now debate whether to allow products inside the sin or ln without parentheses like sin 𝑥𝑦, or if that’s gonna mean (sin 𝑥)𝑦 instead. Also consider sin 𝑥 sin 𝑦. My personal take on these would be to disallow too confusing notations entirely and always use parentheses to disambiguate in such cases, for example in the last case either with (sin 𝑥)(sin 𝑦) or with sin(𝑥) sin(𝑦). If you think: Nested function application with sin, like interpreting sin 𝑥 sin 𝑦 as sin(𝑥 sin 𝑦), i.e. sin(𝑥 ⋅ sin(𝑦)), wouldn’t make much sense, yeah probably true, but at least with ln it does. The function ln  ln 𝑥 is something you actually need sometimes (I’ve seen it written in the form log  log 𝑛 in computer science [admittedly that’s not base 𝑒 anymore]), and it’s pretty nice to not be in need for all the extra parentheses of ln(ln(𝑥)).

@@steffahn you got a good point. I think here the problem is -as you said- not that it's a different notation, but that it is handwritten. Thus there is no such thing like consistent spacings or no spacing. sin sinxy might be sin x y and without parentheses here, you can't know for sure what is meant. I mean yes, in the context it is mostly clear what is meant, but the fact that it's handwritten makes this topic debatable in my opinion.

You will want to learn differential calculus, integral calculus, and differential equations.

@@angelmendez-rivera351 let g(x) = exp(x)

@VeryEvilPettingZoo that's fair. The algebra checks out. I think it's a good idea to just verify where the rule works. Like f and g need to be differential-able g' != g etc.

There is no direct value to my knowledge. The basic premise of the video is that many first year calculus students would suppose that the derivative of the product of two functions would simply be the product of their derivatives only to find it that there is a bit more to it than that! This video just examines what functions would work for that naive assumption.

It's only referred to as a "rule" by analogy to the actual product rule. The exercise was simply to find functions f and g where the statement holds true, knowing that it does not do so in the general case.

I mean, the equation is satisfied when both f and g are of this form.

С*exp(-2*x) also fits

g(x)=0 satisfies the original problem trivially so is a valid solution. If g=g', then f(x)=0 is a valid solution to the original problem. Neither condition is a restriction of the problem originally posed, they're only restrictions on the specific class of solutions his derivation can find.

Yes, but, by the way, the interesting thing is, that in our country such a "parenthesesless" form of logarithm, along with the other functions like trigonometric, is pretty much common. The only case we use parentheses here is when disambiguation is needed: ln(2x • y + 5z)

this is not new kind of derivative; this holds just for special couples of funcs; it is like a differential equation;

I don’t remember the question but it was from a long time ago 😊

@@maypiatt3766 Great!!

Would i get some free marks if i wrote the answer as: F(x) =0 or g(x) =0 or both

"nice" here means that you assume that your function is sufficiently smooth on the domain of interest so that you can differentiate your function, and you're not worrying about g'(x)≈g(x). Not very nice functions f(x)=Weierstrass function (continuous yet non differentiable). Not differentiable anywhere. f(x)={ 1 if x is odd, -1 if x is even, 0 otherwise}. Not differentiable at those discrete jumps. f(x)=|x| near x=0. Not differentiable at x=0

On the board is the note: assuming everything is "nice".

It just means that what is derived is a source of examples, but not exhaustive. There could be other examples. One example not captured here by the final equation, for instance, is f = g = 0.