I am a 64 year old chemical engineer, still working. Math has always been my strongest subject. I enjoy these videos very much. I feel if you do not understand math that well, you will have lot more difficulty in engineering.
@@mesindetrabajinv666 Look at the occupational handbook for job outlook. Chem Engineering was hard but I've heard aerospace is even harder. My concern is, unless you are exceptional, it may be harder t find a job in that field. I started in Chemistry but switched to Chemical Engineering when I found out that BS ChEs were getting almost same pay as PhD Chemists. Probably more important, do what you enjoy.
You have become my favorite UA-camr. Your teaching style is fun, you aren't afraid to show us your mistakes, and you are just enjoyable to watch. I can tell you genuinely want to teach, not just show off your skills. Keep up the good work!!
Man, your enthusiasm is SO contagious! I stood up here in front of my computer and watched the whole thing straight through, with a couple of pauses to reassure myself why some things worked out the way they did! Thank you!
Same! My attention span is such that I can watch a full movie over the course of several days. I saw this video and thought I'd kill a few minutes before going to make dinner and ended up watching the whole thing. I was rapt!
@@fabianwho9797 you can use it in numerical applications as a second order approximation of the derivative: [f(x+h)-f(x-h)]/(2h) = f'(x) + O(h²) in contrast to [f(x+h)-f(x)]/h = f'(h) + O(h) (for h→0, assuming f∈C²)
I'm still a bit lost as to why he had to go that route. Maybe he said why and I missed it. The symmetrical derivative can be used for a symmetrical function about a non-differentiable point, I guess. But why does he use it for sin(x^2)?
@@fabianwho9797 it is very commonly used for numeric derivatives (Central differencing), especially on i.e. images when computing gradients or laplacians or calculating the slope or the normal of a heightmap, also heavily used to solve grid-based fluid simulations
I'm neither a student nor a teacher. I'm just an old fart engineer who loves math enough to realize I really let myself get rusty on the basics. So, although most of the math I do at work is related to number theory, I do appreciate the refresher, particularly since you tend to tackle problems in ways different than what I would use.
This was super awesome! I took calc 1 in highschool, and I've taken calc 2, Differential Equations, and Linear Algebra in college. I don't have many math credits left to take, so I find these videos awesome for keeping me on my math skills. Thank you!!
Seems all ok but this finding of the second derivative using the symmetric derivative as shown is good only for x≠0, see 26:57 where you multiply and divide for 2x. Would be need complete the proof for x=0, i guess is not hard, pluging x=0 in the initial steps, see the formula is also ok
What amazes me is that I'm 47 yo, finished my highschool long time ago, dropped from university on second year, I don't use calculus in my life at all, and I still manage to understand most of this.
Hey there. Before asking this, just wanted to say I love your videos. Thanks to you I've found out about my interest in math as a hobby, and I can't commend how interesting and satisfying stuff like calculus can be when you understand it enough. With that out of the way, I also wanted to ask you a question, which definition of derivative do you prefer? f(x-h) - f(x) lim ---------------- x->0 h or f(x) - f(h) lim ------------- x->h x - h
You've explored tetration a few times, but can the concept tetration be extended to include non-integers? If we define Tn(x) to be a tetration function such that T3(x) = x^x^x and T5(x) = x^x^x^x^x, then what happens if we plug in say 2.5 for n or even i? Is that even possible?
I researched this a bit and it seems very interesting. For anyone interested, go to Wikipedia>Tetration>Extensions (go to the 'heights' section for n) and search up "Tetration Forum" if you want to see more discussion on tetration.
18:53 Actually, you can use L'Hôpital's as long as you don't derive d/dx[sinx]=cosx from it, and there are other ways to prove the derivative of sine is cosine :)
The only thing I like more than trying to follow all of the steps you take during the video is your sheer joy when the thing is finally done! I can't help smiling as well. Thank you for reminding me that i really like math!)
Chain rule? What is the Chain rule? What happened to Chen lu? Did you mean Chen lu? I miss the old days. You are still a great teacher and always had a wonderful sense of humor.
5:30 and what if the graph is not symmetrical relative to the "angle" point, say if there's a sum or difference of a linear function and an abs function with various angle coefficients. What meaning does symmetrical derivative has, will it be relevant? It doesn't seem obvious
I was able to do it by applying the derivative twice so d/dx (sin(x^2)) = cos(x^2)*2x. Then d/dx (cos(x^2)*2x). using the product rule you get -sin(x^2)*2x*2x + cos(x^2)*2. This simplifies to -4x^2*sin(x^2) + 2*cos(x^2)
Can we use the following definition to calculus the second derivative of sin(x²) ? f''(x) = lim h->0 [ (f(x+2h)-2f(x+h)+f(x)) / (h²) ] I don't know if using this formula would be more easier than the other one in the video, I don't have the energy to try it 😂 And also, I don't know if the formula is right or not but I do a little demonstration to proove it and I think the formula is right... Please tell me 👀 PS : sorry for my english, I'm actually french
Definitely, in fact this might be there regular second derivative not the symmetric one, which might exist even if the function isn’t twice differentiable?
I'm always looking for entertaining videos to watch while eating my food. Although I love maths, I would have never guessed I'll end up with this kind of stuff as best eating videos
My students used to keep a catalog of the outrageous things I said over the year. I think "just give up" and "don't be happy to quickly" satisfy the criteria!. I have one question: Doesn't the "'symmetric derivative" (which looks akin to a central finite difference) lead to f' = 0 for absolute value at the critical point?
It's fun when you are seeing the steps you have to take. Not so much when you're standing at a crossroads. Though I learned earlier today that sometimes you just have to make a decision (a big decision, as Bob Ross would say) and stick with it until you hit a dead end.
17:07 After the 4th line, I know why he factored out the 2sin(x^2) but i don’t necessarily get how. The 2nd term in that question doesn’t have the factor sin(x^2)?? am i just reading it wrong Also, I don’t understand why he uses the symmetric derivative for this function to begin with. I don’t quite know if it’s easier, but the function itself has a derivative with the domain of all real numbers, so wouldn’t it be easier to just take the derivative of that again?? i think he said that the symmetric derivative doesn’t imply the existence of the first derivative but if the first derivative does exist then what’s the point?
It's not the 2nd term in the numerator, but the 3rd term (which is where the "-1" then came from). As he explained at the beginning, if you know the normal derivative exists, then the symmetric derivative also exists, and the two are equal. Since you can choose either representation, evidently the symmetric derivative was easier to work with.
This was endless excruciating pain followed by revelation. Sometimes you have to have faith and keep slogging! Wow! You did this as an 8 year old? Proof of alien visitation! I'm calling the History Channel!!!!!!
do you know that derives as you write do not mean derives in mathematics , it means derives in physics , but to calculate a derive in mathematics can be only count with limit
What the other poster said. He skipped a step in deriving the formula for the second derivative and just gave it to you. But all you do is apply the symmetric derivative once to get the first derivative then apply it again to the first derivative. If you want to see the derivation, search for centered finite differences.
Proving that the second symmetric derivative equals the second derivative, on the assumption that the second derivative exists, is not too bad; the function must be differentiable on a neighbourhood of the point, so you can use L'Hôpital's Rule. That reduces it to the symmetric derivative of the derivative. However, proving that the symmetric derivative equals the derivative, on the assumption that the derivative exists, is harder, because it might not be differentiable anywhere else, so you can't use L'Hôpital's Rule. Instead, you can use the same kind of tricks that you'd use to prove the Chain Rule.
Fine and interesting but a little circular. You need to know the 2nd derivative EXISTS in order to apply the symmetric 2nd derivative. The way of knowing that is... pretty much the chain rule.
Everyone is inspired by this great video, but apparently I'm the only one who heard two cries for help. Twice! you asked for a bigger house to fit a bigger board. Someone get this man a larger house!
I do appreciate it! Thank you. I am surprised at how much I remember. I am also keenly aware that I have forgotten so very much. But thank you for the videos!
wait, does it mean you can just use symmetric derivative as a normal derivative? Or is it only second symmetric derivative = second derivative? Or is it just coincidence?
you can use symmetric derivative definition in place of the regular derivative, they should be the same, except sometimes the symmetric derivative will you give you an answer while the regular derivative does not (when there is a sharp turn in the graph, like f(2) = |x - 2|) quoting wikipedia, "If a function is differentiable (in the usual sense) at a point, then it is also symmetrically differentiable, but the converse is not true" this applies for any nth derivative, not just the 2nd
maybe using the fact that sin(x²)=Im(exp(ix²)) helps as d²sin(x²)/dx² = Im(d²(exp(ix²))/dx²) And with your method, exp(i(x+h)²)+exp(i(x-h)²)-2exp(ix²) = exp(i(x²+h²))[exp(2ixh)+exp(-2ixh)]-2exp(ix²)= 2exp(ix²)[cos(2xh) exp(ih²)-1]. Then, we have [cos(2xh) exp(ih²)-1]/h² = exp(ih²)[cos(2xh)-1]/h²+[exp(ih²)-1]/h². For the first part, we know that lim [cos(2xh)-1]/h² = -4x²*lim [1-cos(2xh)]/(2xh)² = -2x² and lim exp(ih²) = 1. For the second part, lim [exp(ih²)-1]/h² = i (as h²->0 and lim [exp(ih)-1]/h=iexp'(0)=i). Thus, lim [cos(2xh) exp(ih²)-1]/h² = -2x²+i . Therefore, d²(exp(ix²))/dx²=(-4x²+2i)exp(ix²) and finally, by taking the imagenary part, d²sin(x²)/dx² = -4x²sin(x²)+2cos(x²)
@@Yougottacryforthis I want to see if bprp can get exact form like x^x = 2 Normally it’s just numerical estimation but he used Lambert W. Can he use the lambert w or other ways to do the request in exact form?
Ok, i have an issue. If you assume sin(alpha+beta) for sin(A). Youre also implying a relationship between A and B. You might say, well ive got x+h and x-h but they are squared. So if i would have gotten to this point i probably wouldnt have persued this attempt because of the implied relationship whuch you cant see in the limit.
Why aren't we deriving normally? Why do we have to use the limit definition? Wrote the comment right before the end part of the video. Guess we're looking to find the hard way to solve it. I thought there was a catch if we solved it with the normal deriving method.
When I saw that we need to find (sin(x^2))'' I was thinking how is this video this long and then BRP pulled out the limit definition of the derivative.
Thank you for showing us two ways to find the second derivative, but it was unnecessary. The first method is way more efficient and practical. The second method added too much time to this otherwise concise video.
If this one limit isn't crazy enough, then try 100 limits: ua-cam.com/video/TglD4Y6lmQk/v-deo.htmlsi=lR_jfa-gI8FVO7VS
already done sir.
easy 🙅🏼♀
This is the first time I was introduced to symmetric derivatives, as my college professors never taught me about it.
Unfortunately, this problem is unsolvable. Not because there is no solution, but because my board isn't big enough.
Why, it's a one line derivation! If you start sufficiently far to the left. 🙂
@@TomFarrell-p9zas long as the size of your letter approaches 0
I have discovered a truly marvelous demonstration of this proposition however this whiteboard is too small to contain.
Assume you have an infinite whiteboard
@@thexavier666 no it tends to infinity
Bro really used the limit definition, legend status
😆
That is a fact!
Pure madness too 😂😂😂 using the Limit Definition to do derivatives of Trig functions is crazy 😂😂
Even worse to integrate with series
I’m just gonna do a reihman sum with infinite rectangles
Now you have to do an epsilon delta proof of the limit for the ultimate presentation of mathematical rigor.
😂
but before that you need to proof that 1+1=2
This would take 12 hours but would be a good video
Xddddddddddd
@@blackpenredpen ... Why do you laugh? It was not a joke.
It's reassuring that I'm not the only one prone to making calculation errors. Great video!
As an aspiring student in AP Calculus, this video was incredible to see! Awesome content! ❤
Thank you!
You're gonna fail AP Calculus blud
@@Sir_Isaac_Newton_😂😂😂😂
@@Sir_Isaac_Newton_when Newton says it 💀
@@Sir_Isaac_Newton_I just got an A on my semester final it’s really not that hard if you pay attention (assuming good teacher)
I am a 64 year old chemical engineer, still working. Math has always been my strongest subject. I enjoy these videos very much. I feel if you do not understand math that well, you will have lot more difficulty in engineering.
Thank you!!
Can you give me some advice for college and engineering? I’m planning to study Aerospace engineering
and even in many aspects of life
@@mesindetrabajinv666 Look at the occupational handbook for job outlook. Chem Engineering was hard but I've heard aerospace is even harder. My concern is, unless you are exceptional, it may be harder t find a job in that field. I started in Chemistry but switched to Chemical Engineering when I found out that BS ChEs were getting almost same pay as PhD Chemists. Probably more important, do what you enjoy.
Im not so sure. Im not applying so much advanced mathematics again in carrier. Even friends that were more advanced had forgotten linear algebra.
You have become my favorite UA-camr. Your teaching style is fun, you aren't afraid to show us your mistakes, and you are just enjoyable to watch. I can tell you genuinely want to teach, not just show off your skills. Keep up the good work!!
Man, your enthusiasm is SO contagious! I stood up here in front of my computer and watched the whole thing straight through, with a couple of pauses to reassure myself why some things worked out the way they did! Thank you!
Same! My attention span is such that I can watch a full movie over the course of several days. I saw this video and thought I'd kill a few minutes before going to make dinner and ended up watching the whole thing. I was rapt!
I don’t think I was ever taught what a symmetrical derivative is in my calculus classes. Thank you.
Im no expert, but in my judgement it is rarely useful for anything, so most people never hear of it
@@fabianwho9797 you can use it in numerical applications as a second order approximation of the derivative: [f(x+h)-f(x-h)]/(2h) = f'(x) + O(h²) in contrast to [f(x+h)-f(x)]/h = f'(h) + O(h) (for h→0, assuming f∈C²)
I'm still a bit lost as to why he had to go that route. Maybe he said why and I missed it. The symmetrical derivative can be used for a symmetrical function about a non-differentiable point, I guess. But why does he use it for sin(x^2)?
@@EmpyreanLightASMRusing the normal def of deriv i got it in about 7 minutes so i assume it was just for funsies
@@fabianwho9797 it is very commonly used for numeric derivatives (Central differencing), especially on i.e. images when computing gradients or laplacians or calculating the slope or the normal of a heightmap, also heavily used to solve grid-based fluid simulations
I'm neither a student nor a teacher. I'm just an old fart engineer who loves math enough to realize I really let myself get rusty on the basics. So, although most of the math I do at work is related to number theory, I do appreciate the refresher, particularly since you tend to tackle problems in ways different than what I would use.
This was super awesome! I took calc 1 in highschool, and I've taken calc 2, Differential Equations, and Linear Algebra in college. I don't have many math credits left to take, so I find these videos awesome for keeping me on my math skills. Thank you!!
This is better than anything on netflix!
31:38 You can hear the relief in that "Yes!" Congrats on getting a good take!
13:05 I'm a CS major (junior btw) watching this lol
These videos make me feel good lol
Thanks for the great video!
I didn't know about symmetrical derivatives until now!
Your excitement is worth of the challenge!
Always a good day when you upload Steve. I always enjoy your videos where you do proofs like this.
Thank you so much!
For cos(2xh) - 1 you can also do the cos^2(xh) - sin^2(xh) -1 = -2sin^2(xh) to avoid the trig in the denominator
Sinx^2을 두번 미분~
첫번째미분 : 2xcosx^2
두번째미분 : (곱의미분적용)
2cosx^2 - 4x^2sinx^2
Seems all ok but this finding of the second derivative using the symmetric derivative as shown is good only for x≠0, see 26:57 where you multiply and divide for 2x. Would be need complete the proof for x=0, i guess is not hard, pluging x=0 in the initial steps, see the formula is also ok
I'm 15 and in year 11 (grade 10), and i haven't officially been taught calculus yet, but i find these kinds of videos super interesting!
You're in for a treat if you go deep in this channel
Proud of you brother. Keep it going
Calculus will keep you fascinated for the rest of your life - even when you are 80. Keep enjoying it.
im a little younger then you but i underatand very well. I remember I didnt understand them a year ago but I just couldnt stop watching these videos.
same here bro
What amazes me is that I'm 47 yo, finished my highschool long time ago, dropped from university on second year, I don't use calculus in my life at all, and I still manage to understand most of this.
This is math candy. Awesome! ❤
Yo your comment is the thumbnail!
That was excellent!! I was on the edge of my seat!
1:53 this was genuinely so funny I love that lol
Now prove that limit is true by the definition of a limit (epsilon delta)
Hey there.
Before asking this, just wanted to say I love your videos. Thanks to you I've found out about my interest in math as a hobby, and I can't commend how interesting and satisfying stuff like calculus can be when you understand it enough.
With that out of the way, I also wanted to ask you a question,
which definition of derivative do you prefer?
f(x-h) - f(x)
lim ----------------
x->0 h
or
f(x) - f(h)
lim -------------
x->h x - h
Me when I try a calculation and I do it the more complicated way:
You forgot to close the parentheses of the first d/dx at 30:27
31:36 the excitement and happiness 👏😘 such happiness in his face very nice to see good answer sir 👍👏👏
wow, your videos are so consistent that i didn't even notice this was from an hour ago
also 25:30 funny integral sign :)
ikr! I didn't notice until you pointed it out. I love this format
You've explored tetration a few times, but can the concept tetration be extended to include non-integers? If we define Tn(x) to be a tetration function such that T3(x) = x^x^x and T5(x) = x^x^x^x^x, then what happens if we plug in say 2.5 for n or even i? Is that even possible?
I researched this a bit and it seems very interesting. For anyone interested, go to Wikipedia>Tetration>Extensions (go to the 'heights' section for n) and search up "Tetration Forum" if you want to see more discussion on tetration.
18:53 Actually, you can use L'Hôpital's as long as you don't derive d/dx[sinx]=cosx from it, and there are other ways to prove the derivative of sine is cosine :)
The only thing I like more than trying to follow all of the steps you take during the video is your sheer joy when the thing is finally done! I can't help smiling as well. Thank you for reminding me that i really like math!)
Chain rule? What is the Chain rule? What happened to Chen lu? Did you mean Chen lu? I miss the old days. You are still a great teacher and always had a wonderful sense of humor.
14:56 (x+h)^2 = x^2 + y^2 + 2xh, (x-h)^2 = x^2 + y^2 - 2xh, select alpha=x^2+y^2 and beta=2xh and you didn't need the sum to product formula :)
5:30 and what if the graph is not symmetrical relative to the "angle" point, say if there's a sum or difference of a linear function and an abs function with various angle coefficients. What meaning does symmetrical derivative has, will it be relevant? It doesn't seem obvious
Now do ε-δ 👀
I was able to do it by applying the derivative twice so d/dx (sin(x^2)) = cos(x^2)*2x.
Then d/dx (cos(x^2)*2x). using the product rule you get -sin(x^2)*2x*2x + cos(x^2)*2.
This simplifies to -4x^2*sin(x^2) + 2*cos(x^2)
No shit sherlock but that’s not the point of the video
26:55 When both limits needed that 2x•2x/(2x•2x), and that finally connected with the answer given by chen lu, then I smirked a little c:
I think we're going to need a bigger board 🦈
Can we use the following definition to calculus the second derivative of sin(x²) ?
f''(x) =
lim h->0 [ (f(x+2h)-2f(x+h)+f(x)) / (h²) ]
I don't know if using this formula would be more easier than the other one in the video, I don't have the energy to try it 😂
And also, I don't know if the formula is right or not but I do a little demonstration to proove it and I think the formula is right...
Please tell me 👀
PS : sorry for my english, I'm actually french
Definitely, in fact this might be there regular second derivative not the symmetric one, which might exist even if the function isn’t twice differentiable?
This was really beautiful; thank you!
I'm always looking for entertaining videos to watch while eating my food. Although I love maths, I would have never guessed I'll end up with this kind of stuff as best eating videos
31:22 Hahaha I thought that I was the only one who get that excited when I finish a "uncrackable problem" like this one
good job
Hey, Guy, you’re great! While I was following you, frankly, I lost a few times, but, you delivered the ship to the quay, in sane!…
That was brilliant!. As a student I enjoyed every bit of that
My students used to keep a catalog of the outrageous things I said over the year. I think "just give up" and "don't be happy to quickly" satisfy the criteria!. I have one question: Doesn't the "'symmetric derivative" (which looks akin to a central finite difference) lead to f' = 0 for absolute value at the critical point?
i can aprove this worked. 30:08 before this. In my head ❤. I am really happy i can do it.
Love the marker drop at the end! Boom! Marker drop!
im never going to use the product rule and chain rule again because i loved doing it the long way. !!!! Nice proof.
Reminds me of the time when I was asked if there is 2 ways of solving a problem, which one should be used? I replied to use the easier method.
That was a very fun 32 minutes of my life. Honestly, this makes me want to practice doing this kind of stuff on my own just because it looks fun lol.
It's fun when you are seeing the steps you have to take. Not so much when you're standing at a crossroads. Though I learned earlier today that sometimes you just have to make a decision (a big decision, as Bob Ross would say) and stick with it until you hit a dead end.
this derivitive AkA fs(x) used also in Mechanics Element finite or Numerical methode to describe descritisation of grid points in the plan
31:36 happiest man alive 😂
We extremely appreciate your effort
31:37 certified mic drop / pen drop moment
I literally solved the first derivative of this last saturday (by the definition)
for the cos(2xh)-1, you can make it into -2sin(xh)^2 😀
17:07 After the 4th line, I know why he factored out the 2sin(x^2) but i don’t necessarily get how. The 2nd term in that question doesn’t have the factor sin(x^2)?? am i just reading it wrong
Also, I don’t understand why he uses the symmetric derivative for this function to begin with. I don’t quite know if it’s easier, but the function itself has a derivative with the domain of all real numbers, so wouldn’t it be easier to just take the derivative of that again?? i think he said that the symmetric derivative doesn’t imply the existence of the first derivative but if the first derivative does exist then what’s the point?
It's not the 2nd term in the numerator, but the 3rd term (which is where the "-1" then came from).
As he explained at the beginning, if you know the normal derivative exists, then the symmetric derivative also exists, and the two are equal. Since you can choose either representation, evidently the symmetric derivative was easier to work with.
@@jb31842haha i totally just didn’t see that at all, thanks for pointing that out
It's simple. Already the first derivative of this function “jumps” very much. For x> 2pi for example.
if anyone used chain rule and producst rule the answer would be: 2( -x^3 * sin(x^2) + cos(x^2) )
Absolutely beautiful!
This was endless excruciating pain followed by revelation. Sometimes you have to have faith and keep slogging! Wow! You did this as an 8 year old? Proof of alien visitation! I'm calling the History Channel!!!!!!
do you know that derives as you write do not mean derives in mathematics , it means derives in physics , but to calculate a derive in mathematics can be only count with limit
Wait I have a doubt: If we have taken the limit only one time, how is this the second derivative of the function?
that limit finds the symmetric second derivative of the function.
What the other poster said. He skipped a step in deriving the formula for the second derivative and just gave it to you. But all you do is apply the symmetric derivative once to get the first derivative then apply it again to the first derivative. If you want to see the derivation, search for centered finite differences.
Basically you can guarantee both h go to 0 synchronously by MVT
Proving that the second symmetric derivative equals the second derivative, on the assumption that the second derivative exists, is not too bad; the function must be differentiable on a neighbourhood of the point, so you can use L'Hôpital's Rule. That reduces it to the symmetric derivative of the derivative.
However, proving that the symmetric derivative equals the derivative, on the assumption that the derivative exists, is harder, because it might not be differentiable anywhere else, so you can't use L'Hôpital's Rule. Instead, you can use the same kind of tricks that you'd use to prove the Chain Rule.
Fine and interesting but a little circular. You need to know the 2nd derivative EXISTS in order to apply the symmetric 2nd derivative. The way of knowing that is... pretty much the chain rule.
Everyone is inspired by this great video, but apparently I'm the only one who heard two cries for help. Twice! you asked for a bigger house to fit a bigger board. Someone get this man a larger house!
Thanks! I appreciate that. 😃
When teacher asks a simple question for 10 points
I do appreciate it! Thank you. I am surprised at how much I remember. I am also keenly aware that I have forgotten so very much. But thank you for the videos!
wait, does it mean you can just use symmetric derivative as a normal derivative?
Or is it only second symmetric derivative = second derivative?
Or is it just coincidence?
If the function is differentiable in the usual sense, then yes. Symmetric derivative = regular derivative
you can use symmetric derivative definition in place of the regular derivative, they should be the same, except sometimes the symmetric derivative will you give you an answer while the regular derivative does not (when there is a sharp turn in the graph, like f(2) = |x - 2|)
quoting wikipedia, "If a function is differentiable (in the usual sense) at a point, then it is also symmetrically differentiable, but the converse is not true"
this applies for any nth derivative, not just the 2nd
some people enjoy using 2 sticks to start a fire....
Man never used the 2nd derivative by 1st principal…. This is so good
17:00
why can you factor out sin(x^2) if it isn’t in the second term
13:11 "If you're also a calculus teacher, you know what to put on the final exam"
Don't give my professor any ideas 😂
Why second symmetrical derivative equals to regular second derivative?
Bro did derivative calculus using 1st principles. I'm beyond impressed
I can't wait for the follow-up video
It's been a long time since I took calc 1, but I'm now getting flashbacks to a bunch of weird limit definitions
"i'm not gonna prove that limit, I'm just gonna use it"
proceeds to prove that limit
I apreciate it. Thank you. We all apreciate you.
maybe using the fact that sin(x²)=Im(exp(ix²)) helps as d²sin(x²)/dx² = Im(d²(exp(ix²))/dx²)
And with your method, exp(i(x+h)²)+exp(i(x-h)²)-2exp(ix²) = exp(i(x²+h²))[exp(2ixh)+exp(-2ixh)]-2exp(ix²)= 2exp(ix²)[cos(2xh) exp(ih²)-1]. Then, we have [cos(2xh) exp(ih²)-1]/h² = exp(ih²)[cos(2xh)-1]/h²+[exp(ih²)-1]/h². For the first part, we know that lim [cos(2xh)-1]/h² = -4x²*lim [1-cos(2xh)]/(2xh)² = -2x² and lim exp(ih²) = 1. For the second part, lim [exp(ih²)-1]/h² = i (as h²->0 and lim [exp(ih)-1]/h=iexp'(0)=i). Thus, lim [cos(2xh) exp(ih²)-1]/h² = -2x²+i . Therefore, d²(exp(ix²))/dx²=(-4x²+2i)exp(ix²) and finally, by taking the imagenary part, d²sin(x²)/dx² = -4x²sin(x²)+2cos(x²)
How to check if we can integrate a function or not?.
Like e^cosx is non integratable so how do we know that it is non integratable
Omg, grea ttiming, I have an exam in 2 days on calculus.
Blackpenredpen be like: Im going to prove this using Epsilon Delta defintion
I will pass this time 😆
So the regular and symmetric derivatives are totally different things right?
very satisfying!
Your class was catching strays at the end. I'm sure some of them would also appreciate this
-4xsin(x^2)
Try e^x + lnx = 0
Nasty. How do you solve that other than numerical estimation?
@@Yougottacryforthis I want to see if bprp can get exact form like
x^x = 2
Normally it’s just numerical estimation but he used Lambert W. Can he use the lambert w or other ways to do the request in exact form?
Well, masochism is also orientation.
Very cool!
Why not use mvt
Ok, i have an issue. If you assume sin(alpha+beta) for sin(A). Youre also implying a relationship between A and B.
You might say, well ive got x+h and x-h but they are squared. So if i would have gotten to this point i probably wouldnt have persued this attempt because of the implied relationship whuch you cant see in the limit.
Ah you're using it the other way around. Still probably would have been too confused to try further.
Looks like a complete nightmare using the definition of the derivative to solve that instead of using the derivative rules.
appreciate the hard work!
Why aren't we deriving normally? Why do we have to use the limit definition? Wrote the comment right before the end part of the video. Guess we're looking to find the hard way to solve it. I thought there was a catch if we solved it with the normal deriving method.
When I saw that we need to find (sin(x^2))'' I was thinking how is this video this long and then BRP pulled out the limit definition of the derivative.
Thank you for showing us two ways to find the second derivative, but it was unnecessary. The first method is way more efficient and practical. The second method added too much time to this otherwise concise video.