Is there a difference?

[@broccolo.fiolaro]

"If you ever feel negative, just plug absolute value around you and you'll feel positive". He is the Bob Ross of math

[@Invalid571]

Someone partied too hard last night. 😎

[@ninjakawasaki1972]

Went from absolut vodka to absolute value real quick

[@kwirny]

@@ninjakawasaki1972 Haha

[@ShAlAmAnAyA3]

that part hahaha

[@romainedixon4560]

Next time i feel depressed im gonna stand between two sticks #absvalueismostpowerfulthing

[@arnavanand8037]

"We are not going to talk about complex values here" I feel betrayed

[@carsonyt5306]

Actually the derivative of the complex value of ln(x) for x

[@nicholasjuricic3683]

Can anyone answer: does this mean ln x and ln |x| differ by a constant? Why / why not.

[@seiken2106]

@@nicholasjuricic3683 I think neither of those is the difference, maybe it's the domain of the function For ln x: x must be >0 For ln|x| : x can be any number other than 0

[@borg972]

This video is gold. I would mention also the integral of 1/x because why it is ln|x| with absolute value is a very common question!

[@carultch]

If you look at the full picture of logarithms in complex numbers, it explains why. complex log(x) = ln(|x|) + i*(angle(x) + 2*pi*k) where |x| is the magnitude of x, and angle(x) is the angle in radians to x, ccw from the positive real numbers. The k is any arbitrary integer. For positive real values of x, angle(x)=0, and we can keep it simple and let k=0, and see that it is simply equal to ln(x). For negative real values of x, angle(x) = pi. The imaginary part is just a constant of i*pi. Let the arbitrary constant of integration be C - i*pi, and you can see that it salvages the result ln(|x|). It is a legal move to have a different arbitrary constant on both sides of the vertical asymptote, because you can't integrate across a vertical asymptote, when the improper integral on either or both sides is divergent.

[@adilsonfranciscoquissai3888]

The smallest details can change everything. Thanks again for that

[@MrHatoi]

You can also set the absolute value of x to be equal to sqrt(x^2), since the square root is always positive, so the square and square root cancel and preserve the magnitude of the number, but not the sign. Then you can take the derivative of this and get x/sqrt(x^2), which is the same as x/|x|.

[@FGj-xj7rd]

Math is Thug Life 🚬😎

[@fountainovaphilosopher8112]

0:45 "keepin' it real"

[@blackpenredpen]

Ognjen Kovačević definitely

[@xevira]

I know this is an old video, but I wanted to say "When your calculus is so LIT that you need to wear sunglasses to see the board properly."

[@Spectrojamz]

😂

[@marvhartigan3677]

haha well said xevira

[@primthon9596]

I always watch your videos before bed 😂 they are great keep going ❤️

[@MathIguess]

"If you ever feel negative, just take the absolute value of yourself" xD

[@no-po9nv]

I'm in grad school and your channel keeps popping up! You've helped me solve two homework problems! You are too cool for school, bro!:)

[@cpotisch]

Alternate method: d/dx(lnx) = 1/x for x>0, and, by the chain rule, d/dx(ln(-x) = 1/(-x)*(-1) = 1/x for x

[@goliathcleric]

I've been staring at the bar all day, craving a drink. That absolute value joke is the only thing that has made me smile today. #absvalueismostpowerfulthing #absvalueishigherpower

[@JayTemple]

I had hoped you'd mention the corollary: This is why the antiderivative of 1 / x is not merely ln x + C but ln |x| + C.

[@sgjuxta]

I don't know why, but I've never liked using piece-wise functions for working with absolute value. I've always felt it was easier to just replace |x| with √(x^2), and then switch it back at the end.

[@ZipplyZane]

You can put absolute value of x on the bottom, but it just gets a bit more complicated. (1/|x|)(x/|x|) = x/|x|^2 = x/x^2 = 1/x.

[@bloodyadaku]

If you extend this definition into complex numbers, is f'(-2) still -1/2?

[@arnavanand8037]

I think yes

[@ashtonsmith1730]

yes

[@dr.rahulgupta7573]

Excellent presentation of the topics. Thanks a lot.DrRahul Rohtak.India

[@diegofelipe5253]

Man. You are amazing

[@scottalder2374]

Are you hungover?

[@johnhumberstone9674]

Really, you are just getting better and better......

[@blackpenredpen]

John Humberstone thanks

[@caducoelho2221]

I love watching your videos, hahaha #YAY

[@Absilicon]

My guy, rocking his shades I see 😁

[@-a5624]

Thanks for this video :)

[@flowerwithamachinegun2692]

0:46 KEEP IT REAL, MAN

[@Julian-ot8cs]

Can you do a video about the complex version of Ln(x)?

[@prismoid00]

Julian Turner kzbin.info/www/bejne/oWPbaJ5njLWqjqc

[@CharlesPanigeo]

He already has kzbin.info/www/bejne/oWPbaJ5njLWqjqc

[@JamesLewis2]

More generally, as a function on the complex plane, ln(|x|) is not even complex-differentiable (but the derivative of ln(x) is 1/x for nonzero x); however, if the domain is restricted to the reals and the codomain is expanded to the complex numbers, then if x

[@pierreabbat6157]

The complex derivative does not exist on the branch cut, which is usually taken at the negative reals, so it still doesn't exist at -2. However, the discontinuity is removable, except at 0, where it's a pole. Btw, it's πi+ln(-x).

[@BeauBreedlove]

I think the most interesting conclusion to be drawn about this video is that it shows that the integral of dx/x is ln|x|

[@JoshuaHillerup]

So, both are the antiderivative of 1/x, but the one with an absolute value is "better" because it has the same domain?

[@awez_mehtab]

That look is my favourite 😎🔥

[@ousmanelom6274]

With you maths become very easy

[@emiliadaria]

them shades, tho' 😎

[@chrisglosser7318]

Let F(x) = sqrt(x^2)/x * ln (x^2/x). Then F’(x) = 1/x for all x in R and 0

[@Armbrust666]

The derivative of the absolute value function is almost the same as the sign function.

[@jonasalberto1236]

In the derivative of lnx we can also write 2/(|X|+X)

[@thedoctor5036]

Best channel 😻

[@mohammadelsayed5715]

You look so cool 😎 , great video 😍

[@armacham]

7:43 if x is a complex number, then abs(x)/x would not necessarily equal x/abs(x)

[@supermonkygamers3]

To differentiate abs(x), couldn't you just write it as sqrt(x^2) and then use the chain rule?

[@samb443]

William Boitor yes, but only for real x

[@PaddedShaman]

If you do that, you end up with 2x / 2sqrt(x^2), which does return 1 for x>0 and -1 for x

[@teavea10]

Not quite. x/sqrt(x^2) is OK for all x not equal to 0 since the numerator will carry the sign. But sqrt is always non negative, so sqrt(x^2) does not equal x for x

[@michaelbobman7964]

A more instructive thing to try would be to differentiate arctan(nx) and observe the behavior of the derivative in the neighborhood of x=0 as n approaches infinity If you wanted to be even more precise, you could instead take the same function multiplied by 2/π and do all this, but that's probably just being nit-pickey

[@flamingpaper7751]

Can you show a proof for the power rule ( d/dx( x^z ) = zx^(z-1)) where z is any complex number?

[@carultch]

Given: f(x) =x^z Want to show that: f'(z) = z*x^(z - 1) Rewrite f(x) using complex exponential (still base e, but it's a multivalued function): f(x) = e^(log(x)*z) Take the derivative with the chain rule: f'(x) = d/dx e^(log(x)*z) = z*[d/dx log(x)]e^(log(x)*z) For any complex number: log(x) = ln(|x|) + i*(angle(x) + 2*pi*k) d/dx ln(|x|) = 1/x d/dx i*(angle(x) + 2*pi*k) will equal zero, as long as angle(x) is a constant. If x is fixed to the real numbers, then angle(x) will be a constant, until you cross zero. For integer exponents, we already have derivative rules that account for this problem point. For functions of x that have non-integer real exponents, they only have real solutions on the positive side of x=0 anyway, so it isn't really an issue of angle(x) changing. The entire imaginary part of log(x) is a constant, and thus has a derivative of zero. Thus: d/dx log(x) = 1/x, when x is limited to the real numbers Compose with what we did earlier: f'(x) = z/x*e^(log(x)*z) Since e^(log(x)*z) is just x^z, this gives us: f'(x) = z/x * x^z And with exponent properties, this reduces as: f'(x) = z*x^(z - 1) This proves the power rule for real values of x, with any complex exponent. It turns out, that it also works for complex values of x, but that's another proof.

[@user_avadakedavra]

could you explain about y' - y = 0 solution of y ln |y|= x+c |y| = e^(x+c) y = +-ce^x? i really get confused about this absolute value.

[@yrcmurthy8323]

I know how to solve it

[@yrcmurthy8323]

y = C • (e^2x)

[@yrcmurthy8323]

Steps are right till the end, but if you have to remove absolute value, you have to square it. |y| = (e^x) • (e^C) |y| = C • (e^x), since e^c is a constant. Squaring a constant gives another constant. Hence y = C • (e^(2x))

[@yrcmurthy8323]

Sorry for late response, didn't see your comment.

[@alwysrite]

so much to learn !

[@someone229]

Nice sun glasses you've got there😎

[@machobunny1]

Seems to me that if y = abs(x) that y is always positive. Yet at about 6:50 you wrote that the value of y=abs(-x) is -x. Umm, really?

[@bismandeep5266]

so derivative of abs (x) is basically the signum function?¿

[@carultch]

Yes.

[@alkamishra9974]

Can we not also use x/abs(x) because 1/abs(x) * x/abs(x) = x/abs(x)^2 =x/x^2(since x^2 will be the same for both +ved and -ves.) =1/x

[@NickKravitz]

Your future is so bright.. You gotta wear shades. Who says derivative domains cannot be sexy??

[@Truth_101_]

what about when all of the lnx is in abs? like abs(lnx) ???? havent found an answer to it anywhere online. im trying to solve an intgeral from 1/e to e of abs(lnx)dx

[@carultch]

You'd have to construct it as a piecewise function, to account for the cusp at x=1. When x=1, abs(ln(x)) = ln(x) So this means your integral has two intervals. integral -ln(x) dx from 1/e to 1 + integral ln(x) from 1 to 1/e Pull out the negative: -integral ln(x) dx from 1/e to 1 + integral ln(x) from 1 to 1/e Get the general integral of natural log: S ___ D ___ I + ___ln(x) _ 1 - ____1/x __ x x*ln(x) - integral 1 dx = (x - 1)*ln(x) Evaluate from 1/e to 1, and negate it: -[(1 - 1)*ln(1) - ((1/e - 1)*ln(1/e))] = 1 - 1/e evaluate from 1 to e: (e - 1)*ln(e) - (1- 1)*ln(1) = e - 1 Add it up: e - 1 + 1- 1/e = Result: e - 1/e

[@fNktn]

It should be x/|x| because if you go into higher dimensions it'll look the same (grad(|r|)=r/|r|) and wouldn't make sense the other way around.

[@stephenstruble5442]

I have a question. I noticed that you once stated that "The domain of the derivative is either equal to or less than the domain of our original function." However, consider f(x) = (x^2-1)/(x-1) The domain is (-∞, 1)∪(1, ∞). f'(x) = 1 and therefor the domain of f'(x) is (-∞, ∞). So, in this case is appears that the domain of the derivative is actually greater than the domain of the original function, albeit only by an infinitesimally small amount. The original function has a removable singularity. Removable singularities are differentiable, I believe. Could you explain why it appears that I just violated your original argument?

Removable singularities become differentiable if you define a function (called an extension) where the singularity has been removed (the "hole" has been filled in), in that case the new function is both defined and differentiable at that point. The original function will still be neither.

[@robertdeaton4720]

What number to the power of itself is -1? I can't seem to figure it out. I know you can solve it by taking e^W(ln(-1)) where W is the Lambert W function. but i cannot find how to take the W of pi * i.

[@holyshit922]

Difference is in domain if x is subset of real numbers

[@domanicmarcus2176]

Is absolute value of any number over itself equal to 1? My professor said no. So why then do they cancel each other out?

[@carultch]

Only if it starts as positive. If it starts as negative, it equals -1.

[@oracle7858]

If you just think about the graphs, it's pretty obvious there's a difference

[@قوسالمطر-ض2غ]

thanks alot ❤

[@user_avadakedavra]

why do you have to hold that all the time? if i was making videos like this i would get a little mic that i could hang on my shirt or wear on my head thanks for the videos

[@AndDiracisHisProphet]

deal with it

[@nigit7451]

Actually, 1/x-1/x=0

[@kinmanwong1763]

If I didn't say x>0, I better said d/dx(ln x) = (ln x)/(xln x) than d/dx(ln x) = 1/x right? Because the domain of (ln x)/(xln x) is same as (ln x)'s.

[@MathIguess]

Them shades though xD

[@zelda12346]

I don't see why you can't dip into the Complex Realm. Cardano's Formula sometimes dips into and out of the Complex plane even if all three roots are real. The imaginary part of ln x with x < 0 is just a constant. The plot of ln x is just has two levels that are parallel to each other in x-y R2 plane when you throw in an imaginary z axis. And the derivative spat out is in fact a realm number. It goes into a wormhole and spits back out somewhere else in the same universe instead of disappearing completely. One of the more difficult bits I found in Complex Analysis was going from a 2D space to a 4D space. My professor used a domain containing a curve and an image of that curve under a transformation. It worked to a degree, but I think it would have been easier on the class if we had already been plotting complex outputs in a 3D space where the domain was strictly real for years or plotting the real part of an output with a complex domain.

[@carultch]

You can dip into the complex number realm. It just isn't necessary to show the proof he intended to show.

[@shenarat8903]

Hey, I have a question that I haven't been able to solve. It is find an antiderivative for the function g, where g(x) = ln|x-1| / x-1

[@carultch]

Given: g(x) = ln(|x - 1|)/(x - 1) Let G(x) be defined, such that G(x) is a function such that G'(x) = g(x). Set up integration by parts, with the log differentiated and the algebraic function integrated: S _ _ _ D _ _ _ _ _ _ _ I + _ _ _ ln(|x - 1|) _ _ 1/(x - 1) - _ _ _ 1/(x - 1) _ _ ln(|x - 1|) Construct result: ln(|x - 1|)^2 - integral ln(|x - 1|)/(x -1) dx Spot the original integral, and call it I. Equate to itself, and solve for I algebraically. I = ln(|x - 1|)^2 - I 2*I = ln(|x - 1|)^2 I = 1/2*ln(|x - 1|)^2 Conclusion: G(x) = 1/2*ln(|x - 1|)^2 + C

[@yaleng4597]

6:35 Chen Lu

[@blackpenredpen]

Yup!

[@mike4ty4]

Even better, the full indefinite integral of 1/x is NOT ln |x| + C. It's rather a piecewise function, or rather ln |x| plus a piecewise function whose domain has two connected components on which it is constant. In general, the general indefinite integral of any function whose domain is composed of more than one connected component will be equivalent up to adding a piecewise function constant on each connected component. E.g. if you wanted to integrate f(x) := 1/(x^2 - 1), whose domain has three connected components: (-oo, -1), (-1, 1), and (1, oo), you will need a function that is constant on all three: int 1/(x^2 - 1) dx = 1/2 [ln(1 - x) - ln(-x - 1)] + C_1 when x e (-oo, -1) 1/2 [ln(1 - x) - ln(x + 1)] + C_2 when x e (-1, 1) 1/2 [ln(x - 1) - ln(x + 1)] + C_3 when x e (1, oo) **NOT** int 1/(x^2 - 1) dx = 1/2 [ln |1 - x| + ln |x + 1|] + C (WRONG!) Keep that in mind! :)

[@ss_avsmt]

What if we define ln(-1) as a complex number not similar to sqrt(-1). Will we have another branch of Mathematics on that?

[@carultch]

We can and we do, have a definition of ln(-1). In fact, we have a multivalued function called complex log, that does exactly that. In general: log(z) = ln(|z|) + i*(angle(z) + 2*pi*k) where angle(z) is the angle CCW from positive real numbers. Also called arg(z) |z| is the magnitude or modulus of z and k is any arbitrary integer The principal value of this function, is the branch cut where k=0.

[@vuyyurisatyasrinivasarao3140]

Super...

[@safayiakako3416]

Thanks

[@zaidsalameh1]

Hi, i have an interesting question:- What is the Derivative/integral of Log base X of 2??

[@trace8617]

you cannot solve this by implicit, at least i tried and failed. rather, rewrite the log as the change in base formula, then take the derivative. with quotient rule or by power rule if you put log(x) to the -1 power. hope this helps

[@LilyKazami]

First, use logarithmic identity to get everything in terms of ln, getting ln 2/ln x. Since ln 2 is a constant we can factor it out of the derivative and integral. For derivative, it's a simple application of the chain rule to 1/ln x, and we end up with ln 2/x (ln x)^2. The integral is a bit stranger, since there is no way to represent the integration of 1/ln x with a finite amount of elementary functions. There is an extant function li(x) which is defined as the solution to this integral, but its values can only be approximated with our math. The answer would be ln 2 li(x) + c.

[@soumyachandrakar9100]

cool glasses #yay

[@Harlequin314159]

A lesson all PhD students learn: when totally sleep deprived just wear glasses to hide your eyes!

[@thomasq5186]

What if complex values were valid? If f(x)=ln(x) and f'(x)=1/x, would f'(-2)= -1/2 or would it be something complex?

[@dekrain]

@Thomas Q: Well, ln(-z) = ln (-1 * z) = ln(-1) + ln z = i tau/2 + ln z. ln' (-z) = 0 + ln' z = ln' z. OR: ln' (-z) = 1/-z * -1 = 1/z = ln' z

[@nathansauveur6704]

Thomas Q The problem would be that are infinitely many solutions to ln(-2), namely ln(2)+(1+2k)pi*i, with k an whole number. So there would also be an infinite amount of derivatives, which we don't like. Or something like that, not sure though.

[@samb443]

Nightish_one there are infinitely many solutions, but they all have the same derivative, so d/dx ln(x) at -2 is equal to -1/2

[@nathansauveur6704]

Sam M Ahh, could be. Not that I disbelief you or anything, I just don't know enough about this stuff to really think about it. Thanks for sharing your answer though :)

[@medpop459]

what about derivative of abs(abs(x)) ????

[@gregoriousmaths266]

Medpop that’s just equal to abs(x)

[@pedrosso0]

"And of course we are keeping everything real in this video" Why is that 'Of course' If we go to the complex world and differentiate it we find d/dx ln(x) = 1/x

[@tajpa100]

Is the derivative x over abs(x)?

[@Kitulous]

Yes, it's either that, or |x|/x, or sign(x), x≠0. Use whatever you want. Technically, from the definition, the abs is on the bottom, but if it was on the top, the answer would be the same.

[@gustavosedano294]

#Yay

[@oscartroncoso2585]

You should wear shades more!

[@sinom]

Those shades O.o

[@nournote]

8:00 it surely would have been much betterr to write x/|x|

[@Humongastone]

But if he wrote x/|x| instead for the chain rule part, then he would get 1/x^3 as derivative of ln|x|??

[@ZipplyZane]

Garand Chua no, he'd get 1/|x| * x/|x| = x/(|x|^2) = x/(x^2) = 1/x

[@carl4578]

is the derivative of arg(z) a thing?

[@CharlesPanigeo]

Arg(z) does not have a complex derivative as it doesn't satisfy the Cauchy-Riemann conditions. More info: mathworld.wolfram.com/Cauchy-RiemannEquations.html

[@sinom]

I would say: f(x) = arg(x). Df=R/0 f'(x) = 0

[@michaelbobman7964]

Arg z fails to satisfy the cauchy reimann conditions. Magnitude and conjugation also fail the CR conditions, actually, if I'm recalling correctly.

[@mehdialami2481]

those glasses though :o

[@tjtaneja1285]

Mans out here in goggles and shit

[@helloitsme7553]

Can you please put your video back on f'(x) i had the answers : a. [0,5). b. (5,6]. c. It doesnt. c. x=5. c. (2,4) d. [0,2) U (5,6] e. x=4

[@blackpenredpen]

Sorry, I removed the original vid since I made a mistake in there. But I uploaded the edited one. : )

[@trangium]

#keepingitreal

[@TheCarlosaceshigh]

can someone help me: Be f(x) = g(x-sin4x), g' (π/2) = √2/2, f' (π/2)=?..

[@aashsyed1277]

yes sir

[@ahmadabualrub2590]

good

[@admink8662]

😎 yay~

[@vicentepoblete6627]

#YAY ! ! !

[@rifaturrahman5779]

Ah, Is ln|x| differentiable in the first place? I mean it's not continuos, so shouldn't be differentiable

[@carultch]

It's differentiable everywhere, other than the singularity at x=0.

[@shivimish9962]

Shady maths 😎

[@dynamic-eu1794]

I feel like I've seen this vid ages ago

[@LFSDK]

I can’t just take the absolute value of myself.. I’m zero.

[@technicallightingfriend4247]

If you take in the bottom So |x|.|x|=(|x|) ^2=x² Then x/x²=1/x ahh!

[@yahianattouf3454]

يو ار دونكي 😂😂😂

[@olegt962]

#GLASSYAY!!

[@catoyuvone8475]

#YIAY

[@deoxal7947]

Disappointed, wanted him to evaluate it negative values.

[@AdityaBdamnson]

S W A G