Partial derivatives 2 | Multivariable Calculus | Khan Academy

[@julianno7860]

I used to don`t like khan`s video, because he talks a lot about concept. All I wanted is to know how to solve a certain problem and some examples. But as you get to higher level, you will truly appreciate khan`s video because he explains concepts, and sometimes very difficult and complicated concepts very well. And understanding is what get you to the higher level.

[@zlatanibrahimovicisbettert7980]

I agree

[@rajaradi802]

You need to understand the conce

[@integralboi2900]

He’s amazing, but I don’t think he gives enough examples, the more you do maths, the better you are. If you’re only listening without doing much examples, you can forget.

[@shaktikashyap6]

And all i want is conceptual understanding

[@dlinegrey]

I was expecting to get 10/100 from my calculus exam, now I expect 15/100, thank you VERY much!!

[@hasanefe95]

That one is pretty much a dream for me, gj

[@gobberpooper]

omg. I'm in a pre-calc class("Advanced" Math II) where we've been learning about composite functions for two weeks now. Just last Friday that's alll I knew up to. Within less than a week I'm working in the 3d dimension and tutoring BC Calc students on integration. I owe you my entire brain Sal.

[@Goingone2]

You helped me pass calc two and now you're helping me pass calc three. Its all thanks to you that I have done well in my past calc classes. YES! haha

[@Goingone2]

you're so fun to listen to. I've spent my night doing statics and designing an offroad car for the SAE mini baja race in portland. haha I have to pick my battles.

[@GeorgeMaj15]

very helpful! I love watching your lectures before my classes and then already knowing most of everything. Thanks!

[@cbasemaster]

why not u became my lecture while i was degree.. yes.. you open my mind that calculus is so interesting.. but it's too late, im postgraduated right now.. since internet getting faster and i enjoy watching youtube..

[@axel39er]

Awesome tutorial i couldnt grasp this for a while in lectures!

[@T1MHughes]

Great videos. I've just started watching these Calculus videos. Your Linear Algebra videos are also particularly useful and I would definitely recommend these to anyone who hasn't seen them! Incidentally, at 6:25, you said that cos(y) could be 5 or pi, of which it could be neither.

[@irenefu7187]

Thankyou so much. Eversince I've started watching your videos, my uni lectures have become redundant. Keep up the work, and I'll definitely be donating ! :)

[@mikeb5174]

ugh haven't taken cal 2 yet but got the waiver to take physical chemistry (basically intro to thermodynamics). this should help a lot. thanks sir! i haven't looked but you should also make videos about integrating calculus to physics concepts. the merger as an idea is still only an infant in my mind.

[@riosaki]

Before i could solve the partial derivative problems but i didn't know how it was like that. Cause i really don't understand what the teacher wrote. = = But when i watched this video i understand where it all came from haha thanks for making such educational vids.

[@Ren520]

thanks. im learning this tomorrow =P i was too lazy to read my book ha cuz u're way better at explaining than some stupid book. THANKS

[@guero303030]

thanks so much for all your great videos. Especially for those like me who need some refreshing.

[@DaOneEnOnly]

@jamball77 you misheard. He says "the cosine of a constant is just a constant". And he writes down cos(y) because its a constant and when taking the derivative of something multiplied by a constant you just throw it out front.

[@rapportplatsen694]

9:06 If everything else is a constant, shouldn't they just be 0 then? If you assume that b = 1 , then the derivates of b^3 should be 1^3 = 3, only be zero. Why didn't you remove the b and c when you did the partial derivates with respect to a?

[@nicolebaar1223]

First, if B=1, B^3 = 1^3 = 1. But if B and C are both constants, say B and C are 1 and 4 respectively, then it would be like saying (a^2)(1^3)(4^1/2) = (a^2)(1)(2) =(a^2)(2) and if you did the derivative of 2a^2 it would be 4a with the power rule. If it was a^2+B+C then it would be like 2a+0+0

[@Toiletbowl93]

Look at this for example, y = 3x. What is dy/dx? dy/dx = 3, right? So same thing, y = kx, dy/dx and what do you get? You get k.

[@DaOneEnOnly]

@nal80lia Hey I'll try to help you out here. y^2 went to zero because the derivative of a constant is zero. However, the the x=a^2*b^3*c^(1/2) the constants don't go to zero because they are multiplied. In the example of taking the partial derivative with respect to "a" the b^3 and c^(1/2) are to be treated like numbers. To further simplify this, lets just say b=2 and c=9. Therefore b^3=8 and c^(1/2)=3. SO 3*8=24. If we substituted that in we would have x=24a^2.

[@aucourant9998]

Excellent video. Thank you.

[@bugfacedog44]

"I'll do it in magenta." You are the man!

[@HouseOfOuroboros]

Well done! Really important stuff and outstanding explanation!

[@Darkarix]

Very nice! The only thing is cos(y) can never be either five or pi. Well done anyway,

[@DaOneEnOnly]

@nal80lia When Khan does it he's doing a general case in which we don't know the values of "b" or "c"

[@mcpunky]

Superb learning material, Sir, best of luck to you on your beautiful mission : )

[@abruptcall]

I'll subscribe~~!!! Thanks!~ Easy to understand the burdonsome concepts.

[@souvikroy4823]

cos(y) = 5 or p i!!!@06:25

[@saunved]

He basically means it could be any number you choose. 5, pi, or 10383892, doesn't matter.

[@hauribest]

+Saunved Mutalik But 5, pi, or 10383892 are all impossible since cos(y) must be within [-1,1].

[@saunved]

+hauribest That's right. I watched it again. What I meant was that "y" could be equal to anything. The instructor made a mistake when he said cos(y) could be anything.

[@hauribest]

Yes I understand what you meant now.

[@sarahbell180]

Actually interesting idea, let y be a complex number and play around with it using Eulers formula, e^ix=cos(x)+isin(x)

[@SunnahTaqwa]

Hardly done calculus before and I've learnt my way to partial derivatives in ONE morning - THANK YOU! Just one question, in the previous video ("Partial derivatives") at 08:55 he says the derivative of a constant such as y^2 is zero; I'm confused about why in this video, at 09:17 (and onwards), constants such as b^3 have remained unchanged in the derivative? Instead of turning into zero? Which would then make the whole derivative zero, i know, weird... but i'm VERY new to this - please help!

[@Taricus]

It's because the constants are multiplied onto the variable, not added or subtracted. They're essentially coefficients when you are holding them constant.

[@staroselskii]

i'm not familiar with the education programme in the us, that's why i was wondering if it was included in the basic school programme or what?

[@sachinrajyaguru3700]

Fantastic explanation. Just a request to make. The frequent use of "I don't know" annoys a lot.

[@Incrue]

@3:15 so this line is inside the plane on the previous video?

[@Bear7747]

I really love these videos.

[@CharacterLimit]

Danielyeh: 3x does not equal dA/dx, but rather the partial (del)A/(del)x. Also, if y=x, then why not just reduce it to one variable? Well, then you don't have a partial anymore. They're different.

[@FierceFang]

DUDE!!! You're awesome.

[@javierzapata6619]

Does anyone know where I can download an app or program like the one he uses to graph the equations?

[@Yashpandey467]

adult site.😈

[@sdklktb]

desmos

[@ThrashAbaddon]

thanks for sharing this video :)

[@Andrei-ds8qv]

great video!! the best!

[@deedubya286]

I've got it! By George I think I've got it!

[@Jjunior130]

these videos are great

[@intdawg]

Awesome stuff.

[@DaOneEnOnly]

@T1MHughes he means instead of. Not that cos(y) could ever equal either of those...

[@tanvirkaisar7245]

great work sire!! :)

[@nicolasoli456]

THANK YOU!!!

[@zyrohnmng]

How can I do the partial derivative where i found the equation that passes through (0,0,0) and (x,y,z) or in my case (w,n,x)

[@danielyeh]

hey i have a question. if if A = x^2 + xy and x = y what is dA/dx? depending when you substitute you can get 3x or 4x. can anyone explain why?

[@Calebkipkurui]

Thanks

[@olsome2]

what is the link to download the 3d graphic?

[@michaelndavid]

They will have the same slope but one tangent line will be parallel with the x axis and the other tangent line will be parallel with the y axis. So no I don't think the tangent lines would be the same line.

[@mikeb5174]

ok. so to my very limited understanding of calculus, the term dx means differential of x, or the infinitesimally small difference of x. what does the partial differential of x mean? is my question even valid?

[@osx101]

love it. not just like it

[@nal80lia]

how comes when we had z=x2+xy+y2 when we did it with respect to x, the y2 turned into a zero because when u get the derivative of a constant it becomes zero. but when you did x=a2. b3. c1/2 with respext to a....somehow the b cubed and the c to the 1/2 power didnt turn to zero...........could you plese explain.

[@HAWXLEADER]

lol i imagine his desk.... a stack or 1-2 tb hard drives! btw you didnt sound like you are melting this time when you rotated the graph... new pc?

[@polyrhythmatics]

So awesome

[@icegirl15]

WHAT´s THE LINK to plot???

[@felipebogaerts]

How do I calculate the tangent plane?

[@Vault24]

exactly

[@ronimakela8457]

if cos(y) is a constant, why isn't its derivative 0? (according to the rules of partial derivation)

[@davidkippy101]

Roni Mäkelä Think of it like a coefficient rather than a constant. If you take the derivative of 5 with respect to x, you get 0 because 5 is equivalent to 5(x^0), so when you do the power rule, 0 is brought to the front, and the final result is zero. Likewise, if you take the partial derivative of cos(y) with respect to x, you get 0, since cos(y) = cos(y) * x^0. However, if you take the derivative of 5x with respect to x, you get 5, since 5x=5x^1, and the one is brought down to the front and the exponent to 0, making the final result (1)5x^0, or 5. Likewise, taking ths partial derivative with respect to x of cos(y)x gets you cos(y). So there's a difference between differentiating a constant alone amd a variable with a coefficient.

[@mattjenner1994]

what about the first order partial derivative of e^(x+y) anyone, pleasee?

[@salahbenkorichi9187]

This might be late but it could help others. Just apply the rule. dfx = x'.e^(x+y)= 1.e^(x+y), (y here is just a constant number not a variable anymore). Similarly with regards to y you would get dfx = y'.e^(x+y)= 1.e^(x+y), (x here is just a constant number not a variable). This example would clarify it hopefully, say you have f(x,y) = e^(2x - 3y), dfx = 2.e^(2x - 3y) (where your 3y= constant number) dfy = -3.e^(2x - 3y) (where your 2x= constant number) Hope this helps.

[@wolterh6]

@hcipriano perhaps he should put a donate button somewhere or a flattr

[@Jesus_777.2]

did anyone else notice that f(.3,.3) is actually .27 not .7

[@monx]

he meant y could be pi or 5 not cos(y) could be pi or 5

[@rogertiger]

he told that the y could be any number, not de cos

[@poojawolves370]

Fun fact about Sal: Sal loves magenta!!

[@H0oLIgAnSsRo0lMaRrSz]

how is calculus so easy for u?!

[@minhochoi7823]

I love you

[@GreyAlien502]

What do you mean???

[@rotflmaopmpqxyz]

"This cos(y) could be 5, or pi or whatever XD

[@STEPHENSENGINEERING]

What happened to 10:25???

[@UnexptedB]

@hcipriano he doesn't need any money, he made money more than 90% of the youtube community gathered, and he quit his job to do this: help students for free for philanthropic reasons. Carefull I said youtube community not youtube.

[@Mollehertz]

how can .3 squared be 0.9 ? ... That does not make sense. 0.3 x 0.3 = 0.09

[@androideart]

haha, I thought you were squaring it, but I c you were multiplying .2 x 2.. ok haha

[@tangoforthemango]

you're fucking amazing, keep the videos up

[@TheEzzieboy20]

You talk kind of like John Mayer

[@hzesen1221]

@rotflmaopmpqxyz i think he meant y could be 5 or pi or wtever LOL

[@DrunkWhiteGuy]

Thank you a lot! You are a great person

[@Ghaiyst]

reminds me of black and white version "Related Rates" problem.

[@nicolasmafra]

Where is the link of the applet(Java)?