I Couldn't Find this Derivative

[@BriTheMathGuy]

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[@realcirno1750]

Sup

[@DJ-jo4vf]

And now the onus is on us.

[@therealcomplex0]

Thank for sharing such an amazing platform with us. I have already begun using it. It's great!

[@sibs6953]

Just cancel the derivatives

[@jamesshelton3827]

Great video, but honestly it would have been much easier if you had just used the Riemann-Cauchy-Euler-Penrose-Bernstein-Cantor-Gelfand-Sobolev-Caratheodory-Hilbert-Noether-Weierstrass method for finding derivatives.

[@fluffydawg6019]

I really wanna know if that's the name of a real method. (And before anyone starts saying "oH yOu wErE NoT pAYiNg AttEnTioN iN cLaSS?", I'm in 8th grade)

[@seymenkapkiner64]

@@fluffydawg6019 it’s not, he just glued the names of famous mathematicians together

[@fluffydawg6019]

@@seymenkapkiner64 thanks...that was kinda dumb of me to not realise it lol

[@seymenkapkiner64]

@@fluffydawg6019 nah its alright man, math is messing with all of our minds

[@esajpsasipes2822]

some method should be named that

[@angelmendez-rivera351]

Problems such as this one are the reason why real analysis exists, and why mathematical rigor is important. What exactly is going wrong here? Doing logarithmic differentiation is equivalent to writing that f^g = exp[g·ln(f)] and applying the chain rule of differentiation. The problem is that the above equality is just not true in general. Think about this: if f(x) = 0 for some x, then ln[f(x)] is not defined, hence exp{g(x)·ln[f(x)]} is not defined either. We say that exp[g·ln(f)] has a singularity at x. However, as long as g(x) > 0, f(x)^g(x) is perfectly well-defined in that situation, being equal to 0. So (exp[g·ln(f)])(x) is undefined, but (f^g)(x) = 0. Hence, this constitutes a counterexample to the equality. So if you take the derivative of exp[g·ln(f)], you should expect this derivative to have singularities whenever x satisfies f(x) = 0, even though f^g may be differentiable at that point x, because the two functions are just not equal, and so neither are their derivatives. Now, lim (exp[g·ln(f)])'(x) (x -> 3·π/2) = (f^g)'(3·π/2) is true. Why is that? Well, f^g is what you get when you remove the singularities of exp[g·ln(f)], and in a similar fashion, the derivative of f^g is equal to the derivative of exp[g·ln(f)] after having its singularities removed. This is why the concept of removable singularities is important in real analysis. This idea is the reason why we allow ourselves to just write f^g = exp[g·ln(f)] indiscriminately, even though it is an abuse of notation: any problems that arise can just be taken care of by removing singularities.

[@EpicMathTime]

The MVP of youtube math comments strikes again

[@garole]

👍

[@duckymomo7935]

I wonder if there is an 'easier' example to motivate analysis this one wasn't so bad but the algebra can in fact get tricky and messy that it might demotivate students?

[@PunmasterSTP]

That was a nice derivation! I really appreciate all of your content, especially since it is high-quality and free. Keep up the great work!

[@BriTheMathGuy]

Thanks! I’ll do my best 😀

[@PunmasterSTP]

@@BriTheMathGuy In that case, just keep doing what you’re doing.

[@adityaekbote8498]

Differentiation* lol

[@PunmasterSTP]

@@adityaekbote8498 Yep, that word should work too.

[@MavrosMinotavros69]

@@PunmasterSTP They're not the same

[@williamnathanael412]

Notice that at 2:56 "this inequality holds true for small nonzero numbers" is only true when the exponent is greater than one. And this is guaranteed by 3pi >2.

[@SoloRenegade]

I like doing the long form of the derivative (fundamental theorem) from time to time, just to practice. Nice to know it may even be necessary sometimes.

[@SabaSa6a]

Looking at the function graph, I realize that there must be a lot of zero terms of taylor series at x=1.5pi. I've check by wolfram alpha that the 1st order to the 9th order derivatives are 0 but the 10th order derivative is infinity. So its taylor expansion at this point does not exist. Very interesting point.

[@maxthexpfarmer3957]

I don’t know if Wolfram|Alpha would necessarily be accurate in such an extreme situation.

[@tamazimuqeria6496]

I've got something like this on my final exam in calc 2 as a bonus and i was so close. Thanks man i learn lot from you

[@BriTheMathGuy]

Nice work!

[@smh1566]

Y'all are high Just find f'(x) d/dx (x^n) = nx^n-1 d/dx (sinx+1)^x = x(cosx)^x-1 f'(3π/2) = 3π/2(Cos3π/2)^3π/2 - 1 Since Cos 3π/2 is 0 f'(3π/2) = 0

[@Tryha4d]

​@@smh1566I don't think you can use power rule for variables

[@dqrksun]

1:17 About the ln(0), derivatives are all about limit. So is it reasonable to treat the lnx as a limit of x going to 0? You will get the same answer as well.

[@HelloWorld-dq5pn]

Thought about that as well.

[@lukandrate9866]

Yes, I also think we can just take the limit as x → 3π/2 of f'(x) as we do with integrals when one of the bounds gives you an undetermined form we just take a limit instead of evaluating

[@angelmendez-rivera351]

@@lukandrate9866 Yes, but that distinction is not something you can brush off. Technically, f'(3·π/2) is undefined, even though lim f'(x) (x -> 3·π/2) exists. What we can say is that, if g represents f' with its removable singularities removed, then g(3·π/2) is well-defined, as is equal to what you expect it to be.

[@lukandrate9866]

@@angelmendez-rivera351 You are probably right

[@stevenfallinge7149]

Only if you know ahead of time that the derivative exists and is continuous at that point. That's the assumption you make by taking the limit of a derivative.

[@joshuanugentfitnessjourney3342]

This video was perfect thank you. As a math student about to graduate I'm getting rusty on the easy to moderate math like calculus. So this helped alot

[@BriTheMathGuy]

Great to hear! Thanks for watching!

[@thephysicistcuber175]

Actually by L'hopital I'm pretty sure you can just take the limit of the derivative as x->3pi/2, as long as the limit exists.

[@lightspeed2014]

It tends towards negative infinity

[@ExplosiveBrohoof]

@@lightspeed2014 No it doesn't, it tends towards 0. ln(sin(x) +1) + (x cos(x))/(sin(x) + 1) tends towards negative infinity, but (sin(x) + 1)^x tends towards 0, and 0*(-infty) is an indeterminant.

[@lightspeed2014]

@@ExplosiveBrohoof sinx+1 tends towards 0 as x --> to 3pi/2, which means e must be raised to higher and higher negative powers to obtain that tiny number, so how wouldn't it tend to negative infinity, I'm new to calculus.

[@lightspeed2014]

@@ExplosiveBrohoof you can even plot it on desmos, IT DOES TEND TOWARDS negative infinity,

[@ExplosiveBrohoof]

@@lightspeed2014 You are right, which is why ln(sin(x) + 1 + [term that approaches 0]) tends towards negative infinity. But this term is being multiplied by (sin(x) + 1)^x, which tends towards 0. As an easier analogy, suppose I have (1/x^2)*(-x), and I want to see its behavior as x-->infinity. The -x term tends towards -infinity, but that isn't enough to prove that the entire function tends towards -infinity, because (1/x^2) tends to zero. It turns out that the whole function tends to zero, even though one of the terms explodes to infinity.

[@EpicMathTime]

Stop doing logarithmic differentiation. Apply the power rule and exponential rule and you can just write out the derivative: f'(x) = xcos(x)(sin(x) + 1)^(x-1) + ln(sin(x) + 1)(sin(x) + 1)^x

[@SjS_blue]

agree, this is what i intuitively went to, not much to think about. it is a nice looking function though

[@pedrosso0]

same

[@x714n0____]

He exchanged six for half a dozen...

[@atharvadinkar8847]

How do you evaluate this at x=3pi/2?

[@zoranivanic3543]

The name of the video is I Couldn't Find this Derivative, but the main problem is evaluation at a certain value.

[@francescomanuguerra825]

Have you searched under the bed?

[@BriTheMathGuy]

Nah, that’s where I keep my integrals

[@aashsyed1277]

@@BriTheMathGuy sure?

[@rzno3414]

@@aashsyed1277 yes

[@anshumanagrawal346]

0:03 Me who evaluated the derivative without a problem: I'm 4 parallel universes ahead of you

[@insouciantFox]

1-cos h = 2sin² ½h Lim[2sin² ½h]^(3π/2)=0 No squeeze required

[@maverickgames5972]

There is an alternative way that doesnt require the clever inequality at the end. we can simplify (1-cosh) in another way Note that cosh = cos(h/2+h/2) = cos^2(h/2)-sin^2(h/2)=1-2sin^2(h/2) Therefore (1-cosh)=2sin^2(h/2) Substitute it to the limit (The following expression omit the limit sign cos it will be tedious to repeatedly type in the comments) (1/h)(1-cosh)^(3pi/2+h) =(1/h)(2sin^2(h/2))^((3pi+2h)/2) =2^((3pi+2h)/2)[sin(h/2)/(h/2)](1/2)(sin(h/2))^(3pi+2h-1) Note that limit h-> 0 sin(h/2)/(h/2) = 1 Thus = (2^((3pi+2h)/2))(1/2)*(sin(h/2))^(3pi+2h-1) Note that as limit h->0 sin(h/2) = 0 Therefore the total result = 0

[@maverickgames5972]

God i hope youtube comment can upload images

[@hickek6607]

You can solve this in 30 seconds using the chain rule: d/dx (sin(x)+1)^x = cos(x)*(sin(x)+1)^x = cos(3pi/2)*(sin(3pi/2)+1)^(3pi/2) = 0*0 = 0

[@td866]

I think you can solve this in 30 seconds with the chain rule but your formula is incorrect. The derivative of the outside is one and the chain rule that follows is to keep "the inside" (i.e., sinx + 1) and multiply the derivative of the inside (i.e., cosx). Therefore, f'(x)= cosx(sinx +1). There is no exponent.

[@YRO.]

Can't you just do it this way? d/dx (sinx+1)^x = x(cosx)^x-1 f'(3π/2) = 3π/2(Cos3π/2)^3π/2 - 1 Since Cos 3π/2 is 0 f'(3π/2) = 0

[@skylardeslypere9909]

To make the squeeze theorem part rigorous, we could say: For any given ε>0, choose δ>0 such that |cos(h)-1|/|h|

[@alexdotdash7731]

Awesome video, so easy to follow!

[@jongyon7192p]

the thumbnail should include the derivative. i skipped it in my recommended at first til i realized it wasnt one of those clickbait videos when another videos linked to this

[@moeberry8226]

What ends up going wrong in this situation with logarithmic differentiation is that if you observe f’(3pi/2), this is equivalent to finding the critical point of the function and when Brithemathguy plugs in 3pi/2 into the derivative you get an indeterminate form of 0 times -infinity. This is is one of the rare cases I have seen when using the zero product property, where (Sinx +1)^x is equal to 0 at 3pi/2 however the second more complicated factor goes to -infinity. Therefore you either need real Analysis or L hospitals Rule!

[@normanfrancisco2063]

What an elegant derivative indeed!

[@BriTheMathGuy]

Thanks for watching!

[@mikeaxarlis4508]

f(x) >= f(3π/2) = 0 and f is smooth so by Fermat theorem f'(3π/2)=0.

[@DK-ek9qf]

Спасибо!

[@BriTheMathGuy]

Thanks so much!

[@NowNormal]

1:22 There's a division by 0 too

[@aniruddhvasishta8334]

Yesss, you finally got sponsored! You totally deserve it man!

[@jontheeditor3122]

Oh, I'm sure he's been getting requests for a long time. He's probably just picky.

[@BriTheMathGuy]

Thanks!!

[@paulhk2727]

"Even if you're comfortable with calculus" Errr... no?

[@edgarboekdrukker2950]

Wow this is not at all how we were taught to derive this kind of function

[@Jack_Callcott_AU]

Good stuff Bri

[@unkownuser4401]

I didn't understand a single thing, but i will learn this soon

[@manucitomx]

Thank you, this was great!

[@BriTheMathGuy]

Glad you enjoyed it!

[@SlimThrull]

3:43 That seems like dark magic to me.

[@yakov9ify]

One can just use the fact that as a composition of smooth functions this function is smooth and thus it's derivative is continuous and thus we can use limits to calculate it. The (sin(x)+1) goes to 0 faster than ln(x) diverges and so the entire derivative goes to 0.

[@MrLuigiBean1]

This was super duper clear, thank you!

[@BriTheMathGuy]

Glad it was helpful!

[@mathpie6851]

I couldn't find this derivative.-------> Now I can.

[@justinpark939]

If I had this on my test, I would question why I did the course

[@BriTheMathGuy]

🤔

[@justinpark939]

Well, it could happen, its a competition more than an exam but if this came up, like only a few people at my year level in the country have a chance at this problem.

[@yat_ii]

0:18 Why not just use chain rule

[@ethannguyen2754]

You could also distribute the [sin(x) +1]^x term and take the limit as x goes to 3pi/2, but that isn’t quite as rigorous since you would have to assume that the derivative is continuous at 3pi/2.

[@jimmykitty]

Interesting one... Such a cool experience!!! Thanks for sharing ❤❤❤

[@mathevengers1131]

Hello. Just saw your features channels list. Thanks for putting my channel in it. And I didn't know that numberphile has another channel.

[@aashsyed1277]

Hi

[@aashsyed1277]

@@mathevengers1131 bumberphile 2

[@aashsyed1277]

Thanks for recommending my channel!

[@jimmykitty]

@@mathevengers1131 Hey, are you thanking me ☺☺?!!! I think that's just my subscription list, wasn't that 😅? I'd subscribed your channel cause there I found many interesting videos on Math... and I'm a Math Enthusiast like you!! 😉❤

[@aashsyed1277]

Thanks a lot! Appreciate a lot!

[@BriTheMathGuy]

You bet!

[@NowNormal]

0:04 Is this (sin(x)+1)^x or (sin(x+1))^x?

[@pedrosso0]

Why will using logarithmic differentiation give us an undefined value if our derivative is defined?

[@hydrostrikehd4661]

Because at 3π/2 we have f(x)=0. The problem is we can't take ln(f(x)) if f(x)=0, so logarithmic differentiation can't be applied. However, i think you can still use logarithmic differentiation and then just take the limit for x->3π/2 and you would still get the right answer (i'm not sure of this but i guess it's probably valid).

[@pedrosso0]

@@hydrostrikehd4661 Thanks

[@sethdon1100]

Turns out that this is just a ruse on indeterminate forms…

[@kinshuksinghania4289]

Mathematics!! Never ceases to amaze!!

[@liamwelsh5565]

Anything that requries a lot of mathematical reasoning = bonus question on test/exam = I don't care

[@karanmeharchandanib2265]

Good one ....nice trick 👍🏻👍🏻

[@BriTheMathGuy]

Thank you! Cheers!

[@adlerdoesstuff1872]

Ignorance Is Bliss

[@mformathamatrices9801]

I thought you were going to say about weierstrass function ,lol😂😂

[@aashsyed1277]

Haha

[@BriTheMathGuy]

😬

[@charlesdecarlucci3065]

The plot should not have an asymptote at about 2 pi. The value of f(x) is 1 at 2 pi.

[@anshumanagrawal346]

Ok, I know this is probably not a great place to ask this, but I recently I came up with something that might be true, I've not found any counter examples, and I suspect there is probably a proof for it using multi variable calculus or something. So, let me start at the beginning, when evaluating the derivative of functions of the form f^g, I was taught as a shortcut/trick to evaluate it by treating each function as a constant and the other as the variable one by one and adding the result from both, my teacher told me this is just a trick to remember the result that you would otherwise get by logarithmic differentiation, and I proved for myself later on and I was like ok this is a thing that works. But recently I was came across partial derivatives in Physics and co-ordinate geometry, but I wasn't taught what it was, just as a trick to solve questions. After some googling and not understanding what it is, I let it go. But after sometime, I realised that the rule for differentiation of functions of the form f^g can expressed as the sum of the partial derivates of this function with respect to both f and g. And soon I realised that this exact thing is exactly what the addition and product rule for the derivative are (or atleast it is consistent with them), I researched about it about a but didn't find any answer, however I came across something regarding multi-variable calculus and how the partial derivatives are based on that. I also, saw a MSE Post about this(The differentiation of the f^g type functions) and someone said that we can use a "rule" of derivatives that says if a function depends on two "variables" (Or really function), it's derivative is just the sum of rates of change both the variables around that point. So, to conclude it, my question is, is it true if we have a function which depends on two (or potentially more) "functions" (Or maybe you can say variables), then it's derivative wrt to 'x' (where the variables are functions of x) is the same as the sum of the partial derivatives wrt each of the variables?

[@sekharapramod7819]

As far as I know, yes. The total derivative of a function is equal to the sum of its partial derivatives.

[@anshumanagrawal346]

@@sekharapramod7819 ok thanks bro

[@parlor3115]

Aren't all the numbers that can be written as 3π/2+2kπ problem points?

[@frentz7]

yea logarithmic differentiation with first step " ln (y) " is not so great for functions y that sometimes equal 0.

[@jonathandawson3091]

I mean, du^v = u^v ln u dv + v u^(v-1) du using the chain rule. The rest is simple.

[@bengtbengt3850]

Very nice! Also, when you have (1-cos(h))/h, you could use the identity 1-cos(h) = sin^2(h/2) and that sin(h)/h -> 1 as h -> 0.

[@olahalyn4139]

Sorry I am a bit confused. At the point where we find f'(x) = ((sin(x) + 1)^x) * [x * cos(x) / (sin(x) + 1) + ln(sin(x) + 1)]. Considering (sin(x) + 1)^x = 0^x = 0. Does the fact that ln(0) matter. Isn't 0 multiplied by an undefined number still 0?

[@xtremeblaze777]

Um can’t I just use chain rule here? Looking at this function, at some point I will have to multiply by the derivative of the inside: (sinx + 1)’ = cosx. Then cos(3pi/2)=0 and since this term is multiplied by everything else, the entire quantity is 0.

[@ishaangunjan25yearsago42]

Brilliant but ykw Imma stick to d(3/2π)/Dx = 0 because the function is a constant.

[@pashimanu0073]

But the first one also had (sinx + 1)^x multiplied so at 3π/2 it also gives zero. What's the problem if rest of the thing is not defined

[@Anujkumar-my1wi]

I was wondering why the value of the function at x=a i.e f(a) is not considered in the definition of limit ? when taking limit as x approaches a.

[@tgx3529]

Interesing solution. I used two limits First Is (- x* log(1/(1+sinx)))/x* exp(x*log(1/(1+sinx)))=0 Second x*cosx/[(1+sinx)*x*exp(x*log(1/(1+sinx))))=lim x*cosx*(1+sinx)^(x-1)=0

[@quentinrenon9876]

couldn't you also use a limit to evaluate your derivative ? Just distribute the parenthesis, second term goes to 0 and the first needs a little limit work but also goes to 0

[@kairostimeYT]

L'Hôpital's Rule makes this question a joke.

[@kairostimeYT]

I like the method presented in this video more. L'Hôpital's Rule is kind of boring after all.

[@YRO.]

@@kairostimeYT l'hospital rule is cool af

[@yoav613]

Nice,but you could also find the limit in the first expression that you get and get 0 too

[@angelmendez-rivera351]

Yeah, but taking the limit is not technically correct in that situation.

[@yoav613]

@@angelmendez-rivera351 give me an example that using the limit do not give the correct answer

[@yoav613]

@@angelmendez-rivera351 at least here taking the limit at 3pi/2 gives 0

[@angelmendez-rivera351]

@@yoav613 The limit is indeed equal to 0, but that the limit is equal to 0 does not imply the function itself is equal to 0 at that point.

[@yoav613]

@@angelmendez-rivera351 what is the derivative of x^2lnx+x at x=0?

[@gedlerarevalo9064]

Couldn’t you just “e” both sides that way you can eliminate the ln.

[@suman-rw2zg]

Thanks for this video , i am from India , lots of love from India

[@BriTheMathGuy]

Thanks for watching!

[@depressedguy9467]

It's actually without definition possible

[@BriTheMathGuy]

Cool!

[@aybaws]

All that for 0

[@user-jm6gp2qc8x]

0. Bruh, always remember that you can take the derivative of anything. But integrating, that's hard.

[@BriTheMathGuy]

🤔

[@FireSwordOfMagic]

The Weierstrass function would like to have a word with you

[@anantajitmukherjee81]

I feel this function is non differentiable

[@lacasadeacero]

its 0, cuz f(3pi/2)=0 and its growing up for x after 2pi/2 so its maximal

[@blockthrower3947]

couldve also just taken the limit of the function itself couldn't? basicly limit of t*ln(t) as t goes to 0 is 0 and not much more

[@anantajitmukherjee81]

At 3pi/2

[@AmirHX]

Only trick for solution 😉

[@neuralwarp]

But the thumbnail said d/dx .. aren't you going to find that?

[@magnificentworld324]

Can't 3π/2 be the answer

[@user-jm6gp2qc8x]

Apply log on bts, take derivative bts. Simple no?

[@finmat95]

I don't get it

[@rubenvela44]

Sorry it's just with out you i'm a nobody

[@BriTheMathGuy]

Without supporters like you I’m a nobody!

[@joshmcdouglas1720]

I got to the same point as you when you had lim h->0 ((-cos(h)+1)^(3pi/2+h)/h) but here I used L’Hopital’s rule, and yes it was a super long mess differentiating the numerator but in the end it was only 2 terms (by the product rule) and both of them had a (-cos(h)+1) somewhere in them which made both of them nicely cancel and just be left with 0 for the answer.

[@mihaleben6051]

d/dx is easy, derivatives arent

[@lukeskywalker691]

No dislikes Noice

[@namanmakhija4692]

This video was really good But I did it much faster using short cut method As Derivative of a^x = a^x * log(a) and then putting the value of x

[@Danitux11]

you overcomplicated a lot. i tried to do it before you and i got it much easier. here is what i did: f(x) = (sin(x) + 1)^x we divide it in 2 functions g(x) = sin(x) + 1 s(z) = z^x in s(z) "x" is not an input so we just make another variable ok so now we use the chain rule (derivative of a composition of functions is the derivative of the outer function evaluated at the inner function times the derivative of the inner function. also the composition of f(x) in this case is s(g(x)) derivative of s: z^x times ln(z). but we have to enter the input of g(x) so we substitute g(x) as z: (sin(x) + 1)^x times ln(sin(x) + 1) and now we multiply it by the derivative of g(x) (it is cos(x)): (sin(x) + 1)^x times ln(sin(x) + 1) times cos(x) now it seems so complicated but if we do 3pi/2 on the first sin(x) we get (-1 + 1)^x times ... the -1 +1 equals 0 so then everythings just cancels. is there something i did wrong or you didnt notice this method?

[@spiceymibba8078]

Ooof

[@BriTheMathGuy]

😬

[@anantajitmukherjee81]

Let's see😅

[@siquod]

Way too complicated. Logarithmic differentiation is too complicated as well, apart from not even working here. If you know how the multivariate chain rule works, you can solve this directly in your head in less than half a minute. ((sin x + 1)ˣ)' = (cos x)ˣ + x (sin x + 1)ˣ⁻¹ = 0ˣ + x·(-1+1)ˣ⁻¹ = 0 + 0 = 0

[@kilian8250]

That derivative is incorrect though

[@kilian8250]

Your ”derivative” just happens to equal the actual derivative at x=3pi/2.

[@siquod]

​@@kilian8250 You're right, I seem to have forgotten to factor in the differential of the power operator, which in this case diverges, so the method doesn't work here either. What an embarrassing mistake. I downvote my own comment.

[@YRO.]

You can do it this way d/dx (sinx+1)^x = x(cosx)^x-1 f'(3π/2) = 3π/2(Cos3π/2)^3π/2 - 1 Since Cos 3π/2 is 0 f'(3π/2) = 0

[@siquod]

@@YRO. Unless I'm missing something highly nontrivial, your derivative is even more wrong than mine. What are your intermediate steps?

[@paperwarri0r446]

1000th like

[@BriTheMathGuy]

Thank you!

[@paperwarri0r446]

@@BriTheMathGuy you are welcome! Thanks for an amazing video!

[@yat_ii]

Wow thats a lot of likes

[@jrcarlyon680]

Stop waving your arms around please it's so annoying

[@kaczXX]

6666th

[@BriTheMathGuy]

😲