Math for fun#16, 2nd derivative rule?

Math for fun, 2nd derivative rule?
Is y'' = (y')^2?
Is f''(x) = (f'(x)) ^2?
Is d2y/dx2=(dy/dx)^2?
fake 2nd derivative?

blackpenredpen

[itamargolomb]

Loved the way that MIT professor treated the question - really seriously

[luizestilo]

Math teacher: dy/dx in *not* a division, this is simply a standard notation, so *do not* treat dx/dy like a division, but as a single function.
Math teacher 5 minutes later: 4:08

[elifc]

I dont know why but your energy makes me smile
You seem very happy

[zelda]

"And this is how you keep up with constant calculations."
I see what you did there.

[mairisberzins]

This channel is amazing. I am a highschool student and i wanted to self taught calculus.
I just opened up red pen black pen and cluelesly watched his vids until slowly and slowly i started to understand things.
First time in my life i've had such experience learning anything other than language. Where you just watch/listen to someone and you somehow start understanding it from scratch

[jamesnotking]

eh actually the first things that pops up my mind is any constant function y=c so that y'=y''=y'''=...=0 lol

[simonshugar]

technically the answer could be - ln | x + c1 | + c2 since | x - a | = | a - x | for any a and constants can be negative

[iyziejane]

It's neat that y' = y^a for any a > 1 leads to a blow up to infinity in finite time, because of the contrast with the exponential in the case a = 1.

[Sharaton]

More precisely, the solutions are on the form
y = {-ln(x - C) + D_1, if x > C
{-ln(C - x) + D_2, if x < C
for constants C, D_1, D_2, since you don't necessarily have the same constant on different sides of x = C.

[Kanapek]

If constant functions work too why you didn't get one as a result?

[kokainum]

One minor remark. All you said is correct ofc, but it's good to know that this formula works locally. So in small interval it has to be either constant or given by this formula, but if we want to have a function with maximal open domain than we can take different C2 for x's on the 'left' side of C1 and different for x's on the 'right' side of C1. Also on the one side of C1 it can be given by this formula and on the other it can be just constant. It's just a remark for people wondering what are all the maximal solutions (the ones for which we cannot increase domain).

[Gr]

Congratulations! Great video, keep going. I like your way of teaching :)

Anyway, I think you must analyze also singular solutions. In fact, just the function y(x)=C (C a constant) is also a solution of the ODE. The reason you "lost" that solution is because you divided to be able to do separate variables integration.

[Pete-Prolly]

I never saw this notation in class, we used y'' or ƒ''(x) & on my 1st Calc test I saw this:
Find d²y/dx²|ₓ₌₂
🤦🏻‍♂️ (there were functions to solve for)

[wabc]

I realize the reason we don't get any constant function as an answer, despite not setting any limits on the type of function, is that we assumed we were not dividing by zero.
If we had a constant function, z = dy/dx = dy = d^2 y = dz = 0. So at 4:25 we would be dividing by 0.

[MGSchmahl]

You could have saved a couple of steps by noting that |-x+c_5| = |x+c_6|.

[wabc]

10:17 "We cannot combine them anymore"
y = -ln|-x + c_4| -ln|c_6| where -ln|c_6| = c_5, or |c_6| = e^-c_5
Then y = -ln|ax + b| where a and b are constants (a = -c_6 and b = c_4 c_6)

[krisbrandenberger]

So in general, for every integer n>= 2, the nth derivative of y with respect to x is equal to the nth power of the first derivative of y with respect to x whenever y is a constant function.

[mamadetaslimtorabally]

Just amazingly genius!!! You are a sport !

[varshinilolla]

It's sort of having an integration 101 for me, without ever doing it myself! This never seemed like the conventional calculus class you'd have after your secondary schooling.

Cheerio, elegant explanation!

[armacham]

I show there are additional solutions, namely y = -ln(ax + b) + c, where a, b, c are real numbers

For example, y = -ln(2x) also works

I guess having absolute value inside the ln is optional, unless you care about the domain for which it is true that f'' = (f')^2

for example, if your function is y = -ln(abs(x)), then it's not always true that f'' = (f')^2. It's not true when x = 0, because there is no valid first- or second derivative when x=0.

Also: if a = 0, then -ln(b) + c is a constant, so this formula includes the trivial case where f(x) = c, the same cannot be said of the answer given in this video, which omits the constant case