Can you do a follow-up explaining why you didn't simplify further by cancelling a (dx/dy)/(dx/dy) out of the answer to get (d/dy)/(dx/dy)^2? It's been 20 years since I actively practiced calculus, so I'm a bit forgetful, but I love watching your vids!
The first term in the final numerator, -d^3x/dy^3, should be -(dx/dy)(d^3x/dy^3). The mistake was in the line before factoring out d^2x/dy^2. A good way of spotting this sort of mistake at the end is checking for dimensional homogeneity. The second term in the numerator has dimensions of x^2/y^4, so the first term also needs to have these dimensions.
Missed a dx/dy on the num since you collected the (dx/dy)² from (dx/dy)³, making the first term on num should be (dx/dy)² [ (dx/dy)(-d³x/dy³) + 3(d²x/dy²)² ]
Isn't it possible that d³y/dx³ could be 1/(dx³/d³y) simply because dx³/d³y may not have been defined to be anything else yet (to my knowledge), so we can make it as such? I guess it depends on what the operator 1/d could suggest. d and ∫ are inverses. ∫dx=x, ∫dσ=σ, etc. Sure, they're multivalued functions, with infinitely many branches from all the constants we could add, but for simplicity we can focus on just one branch. So 1/d would undo 1/∫? There might have been a joke Dr. Peyam video using something like this, but I don't know of any too-rigourous maths that's used it.
Can you translate the videos into Arabic, because I am an Arab and I love hearing you very much. I want you to ask a question: Is this a method of study? ❤❤❤❤❤❤❤❤
Yeah , but remember only in (dx/dy)³ not in dx³/d³y .. Also I think it's just how this is normally taught in high school I guess.. It's easier to remember..
if 0.9999.... is 1 ( since there is no real number between them ) then is [0.999...] =1? or is 1 > 0.9999... by dx which will result in [0.999...] = 0 i am confused🙏
@@theweirdwolf1877Technically 0.999..... is another way of defining 1 , like how you define 1/3= 0.3... It's because there's is no real number bigger than 0.99999.. but less than 1 , suppose if you stop at 0.9999999 it's finite number not infinite 0.999.... You can also prove it this way 1) X=0.999... 2)10x=9.99.. 2)-1) 10x-x=0.99..-0.999 ... 9x=9 X=1 This is correct algebraic method.. There is also other methods as well.. So [0.999...] should be one... . I don't care about your math teacher , search up this on Google if you don't believe..
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Why stop there? If you keep going via induction, you get the subfactorial to show up.
f^(n)(x) = (-1)^(n + 1) * n!! * (d^2x/dy^2)/(dx/dy)^(2n - 1)
This is why it’s important to remember that dy/dx is not a fraction.
You can treat it like a fraction though
just remember the dx in the bottom is a single variable, and can't be split into d and x
@sepdronseptadron tbh how I would say is that dy/dx is a quotient. But yeah that's true.
dy/dx=d(y)*(1/dx) it is fraction now
@@caniraso *PEN WHACK*
Thank You, now I know hahaha
Can you do a follow-up explaining why you didn't simplify further by cancelling a (dx/dy)/(dx/dy) out of the answer to get (d/dy)/(dx/dy)^2? It's been 20 years since I actively practiced calculus, so I'm a bit forgetful, but I love watching your vids!
d^2y/dx^2 is not a fraction. To be accurate, it really means d(dy/dx)/dx, which IS a fraction.
at 6:42 I think you have forgotten that dx/dy was risen to the 3td power, and you collected a 2nd power ;-)
And i thing, by his voice at the end, that this answer did not match his previous answer, but he kept going
6:42?
@@niom9446
Uh oh, I think he edited it out of the video 😳
@@NonTwinBrothers oh…
Thanks for pointing that out! I just trimmed that part of the video out to avoid further confusion.
Missed one dx/dy in the numerator
Hey guys! Don’t be a third derivative!
(Distance is the 0th derivative, velocity is the 1st, acceleration is the 2nd, …)
Jerk.
Jerk
The third derivative is a jerk
@@themathematicstutor4092why are you this rude to it? 😂
@@themathematicstutor4092 Plz "Snap" out of it bro.
It’s a good day when bprp uploads
I think there's a mistake when u took the common factor (dx/dy) it'll be [(-d³x/dy³)(dx/dy)+3(d²x/dy²)²]/(dx/dy)⁵ at 6:42
The first term in the final numerator, -d^3x/dy^3, should be -(dx/dy)(d^3x/dy^3). The mistake was in the line before factoring out d^2x/dy^2.
A good way of spotting this sort of mistake at the end is checking for dimensional homogeneity. The second term in the numerator has dimensions of x^2/y^4, so the first term also needs to have these dimensions.
Missed a dx/dy on the num since you collected the (dx/dy)² from (dx/dy)³, making the first term on num should be (dx/dy)² [ (dx/dy)(-d³x/dy³) + 3(d²x/dy²)² ]
I would like to see sin(cos(x)) = cos(sin(x)) the solutions would probably be wild
No bro there is no solution of this problem
@@AadarshTiwary-en2hy of course there are. Just check WolframAlpha for four different solution sets. They may be complex, but they certainly exist.
@@AadarshTiwary-en2hy hes probably asking about complex solns
i checked wolfram, the solutions are complex and WILD, but they are closed.
not sure about sinh(cosh(z)) = cosh(sinh(z))...
yo i wanna see this too!
In the moment of the evidence of (dx/dy)², in the penultimate line, you missed one dx/dy, that must be multiplied by -d³y/dy³, in the numerator.
Could you please do the integral from 0 to inf of (arctan(x))^2/x^2 and then from 0 to 1? 😊
Fan de Axel Arno ?? 😏
@@sachavalette1437 yep 😂 j'avais envie de voir si il allait approcher le problème d'une autre manière
Isn't it possible that d³y/dx³ could be 1/(dx³/d³y) simply because dx³/d³y may not have been defined to be anything else yet (to my knowledge), so we can make it as such?
I guess it depends on what the operator 1/d could suggest. d and ∫ are inverses. ∫dx=x, ∫dσ=σ, etc. Sure, they're multivalued functions, with infinitely many branches from all the constants we could add, but for simplicity we can focus on just one branch. So 1/d would undo 1/∫? There might have been a joke Dr. Peyam video using something like this, but I don't know of any too-rigourous maths that's used it.
Don't be a d^3x/(dt)^3
Technically, wouldn’t it be correct to say that d²y/dx²=1/(dx²/d²y) ?
@0:55 - Shouldn't that have been f'(x)*dx/dy, since you are taking derivatives with respect to y?
It is a fraction
So after trimming the video, is the final ans correct now?
Hey, are there any health implications for working around so much dried ink? Are there such things as healthy board pens?
yes, you'd become extremely susceptible to dried ink syndrome. Symptoms include being hypnotised to write math on and on and on
Can you translate the videos into Arabic, because I am an Arab and I love hearing you very much. I want you to ask a question: Is this a method of study? ❤❤❤❤❤❤❤❤
I can smell the dry erase marker through the screen
Now do the reverse and you get the solution for the differential equation
Give us some examples
glad that brilliant changed its font😅
t the end isnt it legal to flip the (dx/dy)^3 to the top to make it (dy/dx)^3?
Yeah , but remember only in (dx/dy)³ not in dx³/d³y ..
Also I think it's just how this is normally taught in high school I guess..
It's easier to remember..
@@xninja2369 hm ok
One can understand this by 2nd order differentials
What is the differens between d/dy and dx/dy?
"d/dy" is an is an operator which takes a function of y to its derivative.
"dx/dy" is the derivative of x with respect to y.
@@APaleDot d/dy is when you have to take derivative, dx/dy is the derivative?
@@tychovanbrabant5279
dx/dy is a derivative, yes.
d/dy turns a function into its derivative.
From indai 🇮🇳
People just need to stop using d$variable1/d$variable2 where it’s not denoting an incomplete derivative or the exact f’($variable2)
Inspiring 😮
But the real question here is what is d²x/dy²?
He purposely made the mistake in second last line to rage bait interaction in comments
if 0.9999.... is 1 ( since there is no real number between them ) then is [0.999...] =1?
or
is 1 > 0.9999... by dx which will result in [0.999...] = 0
i am confused🙏
Yeah I was also confused about that
My maths teacher said it was 0 though
@@theweirdwolf1877Technically 0.999..... is another way of defining 1 , like how you define 1/3= 0.3...
It's because there's is no real number bigger than 0.99999.. but less than 1 , suppose if you stop at 0.9999999 it's finite number not infinite 0.999....
You can also prove it this way
1) X=0.999...
2)10x=9.99..
2)-1)
10x-x=0.99..-0.999 ...
9x=9
X=1
This is correct algebraic method..
There is also other methods as well..
So [0.999...] should be one... .
I don't care about your math teacher , search up this on Google if you don't believe..
dx is not a value, it IS an infinitesimally small amount
@@bruhifysbackup the point is that 0.9999… is less than 1 by an infinitesimally small amount, that is, dx
@@bruhifysbackupdx definitely is a value with an infinitesimal order of magnitude. It is very, very small.
🗿
Second last step is wrong?? 😊 when u r taking (dx/dy)^2 as common then bracket must hv first term -dx/dy.d^3x/dy^3
anze yu
Abuse of notation