I differentiated the quadratic formula

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I differentiated the quadratic formula

Рет қаралды 244,884

2 жыл бұрын

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This is my first time differentiating the quadratic formula and I did it wrong first! We can first solve the quadratic equation ax^2+bx+c=0 and will see that x is actually a function of a, b, and c. So when we differentiate the quadratic equation with respect to a, either we have to do partial derivative or we will have to keep in mind that x is a function of a and use implicit differentiation.
I derived the quadratic formula while blindfolded: 👉 • I derived the quadrati...
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Пікірлер: 448

@blackpenredpen 2 жыл бұрын

Visit 👉 brilliant.org/blackpenredpen/ to learn more calculus and algebra (20% off with this link)!

@simonmultiverse6349 2 жыл бұрын

What you need is d(roots) / d(coefficients) which is possible, because there is ONE MORE COEFFICIENT than there are ROOTS. This derivative is actually a matrix of coefficients. If you try d(coefficients) / d(roots) it doesn't work. It might work if you reduce it to a standard form, for example the highest power of x could have the coefficient 1. Then you would be differentiating the coefficients wrt THE SAME NUMBER of roots.

@simonmultiverse6349 2 жыл бұрын

As Shakespeare wrote, 2 b squared or not 2 b squared: that is the question!

@herbie_the_hillbillie_goat 2 жыл бұрын

@@simonmultiverse6349 I think you slighlty misquoted the great mathematician, Shakespear. I believe it was, "2b squared or negative 2b squared: THAT is the question!"

@user-yw9jc1xu3l 2 жыл бұрын

i have a question. can we differentiate that? since ‘ax^2+bx+c=0’ is not an identity but an equation, i think we cant differentiate that expression.

@simonmultiverse6349 2 жыл бұрын

@@user-yw9jc1xu3l Yes, because you differentiate each of the terms. Since it's an equation, the derivative is also an equation. For example if you have P+Q=R+S then the derivative would be dP+dQ = dR+dS

@matniet43 2 жыл бұрын

"It was wrong because it was too easy" A wise monk - some year

@herbie_the_hillbillie_goat 2 жыл бұрын

That's how I always knew my homework was wrong. :D

@suhail9511 2 жыл бұрын

@@herbie_the_hillbillie_goat 😂

@Train_Sounds 10 ай бұрын

@@herbie_the_hillbillie_goat XD

@user-me1sq5ln6b 4 ай бұрын

It's easy if you are not sweating.

@drpeyam 2 жыл бұрын

My multivariable Chen Lu senses are tingling haha

@blackpenredpen 2 жыл бұрын

😆

@frozenmoon998 2 жыл бұрын

Do we have enough Chen Lu and Lu Chen supporters guys? It's Dr P on the horizon!

@arpittttt 2 жыл бұрын

Hahaha 😂

@sergiomarquina2924 2 жыл бұрын

This guy has so much courage ,he could solve such complicated expressions with a smile on his face .

@ImineResveratrol 2 жыл бұрын

because it’s not his first attempt LOL

@robertveith6383 2 жыл бұрын

What other part of his body would he have a smile!?

@RockMetalElectronicPlaylists 2 жыл бұрын

S (black hole) dx ;-)

@jandevries9653 2 жыл бұрын

'complicated', this is high school level maths...

@PwerGuido 2 жыл бұрын

This is not complicated at all for a math bachelor

@kaskilelr3 Жыл бұрын

For those who are confused: In the function: f(x) = ax^2 + bx + c x can take any value (independent of a) Df/da = x^2 In the equation: ax^2 + bx + c = 0 x now refers to the set of solutions x1, x2, which depend on a. Hope it helps

@lampshade6579 10 ай бұрын

This looks like what happens when my alphabet soup has broken bits in it.

@johnchristian5027 10 ай бұрын

Thanks for clearing that up, i was really confused at first!

@warrengibson7898 2 жыл бұрын

Sensitivity analysis is important in engineering because the parameters we use are usually somewhat uncertain. It helps to know what a 5% deviation in a parameter like “a” would do to our answer “x”.

@fizixx 2 жыл бұрын

Interesting

@MarkMcDaniel 2 жыл бұрын

Exactly, that's what analysis is good for.

@panhandlejake6200 10 ай бұрын

Yes, understanding how to optimize a system (usually a combination of equations) can be very insightful -- but also a complex problem. From my limited experience, optimization studies are only performed when optimization is identified as a necessary objective. Instead, I tried to use these types of parametric studies just to understand how systems actually worked in good detail when others were somewhat arbitrarily choosing their design point that reached at least ONE solution. Due to complexity, just choosing can save a great deal of time so long as you are able to reach a satisfactory outcome. When the outcome was not satisfactory, the habit was to just try again - when optimization studies could instead have pointed to a much more educated choice. It has been some time since I was in college and it still seems that this type of approach is only lightly taught.

@YuvrajSingh-en2zd 2 жыл бұрын

Bro, your thinking is just out of the box 📦. You are inspiration for many students. You teach us how to think.

@YuvrajSingh-en2zd 2 жыл бұрын

Bro, from which university you studied ?

@toxhydre8806 2 жыл бұрын

@@YuvrajSingh-en2zd look his previous video, first sentance. :)

@bismajoyosumarto1237 2 жыл бұрын

@@toxhydre8806 Ah yes, in his video titled "you, me and the Stanford math tournament algebra tiebreaker", he said he graduated from UC Berkeley as an undergrad. Thank you so much

@zhelyo_physics 2 жыл бұрын

"Let's do some math for fun" means I drop everything I concentrate on the screen! : )

@blackpenredpen 2 жыл бұрын

Thanks! 😆

@bobbyfreeman7671 2 жыл бұрын

Seeing him not simplify the fraction by taking out a factor of 2 hurt my soul

@j10001 2 жыл бұрын

I feel it, too!! But I suppose he ran out of white board space and only wanted to go as far as needed to match up the two different approaches to the derivative. Still… I have to simplify or I cannot sleep at night! Haha

@MFESPFTWKF 2 жыл бұрын

Its amazing how easy I get distracted from revising my further math exam tomorrow with another math video

@_yuki9953 2 жыл бұрын

Same here hahaha

@mathe.dominio4765 2 жыл бұрын

😅🙏

@roberttelarket4934 2 жыл бұрын

This is novel, unique and ingenious! Absolutely never and would never think of it in this way!

@rezakghazi 2 жыл бұрын

This is an excellent approach that you considered *x* as the function of *a* . But as you know, *x* is also the function of *b* and *c* . So if *x = f(a, b, c)* then we have: The derivative of *x* with respect to *a* as you calculate is: --> *δx/δa = -x^2 / (2ax + b)* So we can calculate also the derivative with respect to *b* as: ---> *δx/δb = -x / (2ax + b)* And with respect to *c* as: ---> *δx/δc = -1 / (2ax + b)*

@garricksl 2 жыл бұрын

That is more like upper-division calculus. It is very hard to grasp at first.

@snoozbuster 2 жыл бұрын

A fascinating pattern emerges: the partial derivatives with respect to x_n (the constant term for x^n) are of the form -x^n / (2ax + b).

@roros2512 2 жыл бұрын

those deltas looks super cool, how you do it?! =0

@justanotherguy469 2 жыл бұрын

@@roros2512 That's what I'm tryna find out! Use to be able to copy and paste from Wikepedia, but can't anymore.

@numbers93 2 жыл бұрын

@@roros2512 those are dirac deltas though? they’re not usually meant to denote partial derivatives….

@hydropage2855 2 жыл бұрын

I would love to see you differentiate the first form with x isolated, but show us how you could keep simplifying it until it becomes the simpler form you got after. Show how they’re both equal, but starting from the complicated one

@christiansmakingmusic777 2 жыл бұрын

The broadest construct views all of the variables as functions of all the others, so b is not constant with respect to a. You can see this world if k of a natural number N as the product of two integers written using the division algorithm as (px+c)(qx+d)=N. The coefficients are going to be related to the particular x as a ring base and Vice versa.

@GamerTheTurtle Жыл бұрын

So if x = f(a, b, c) then we have: The derivative of x with respect to a as you calculate is: --> δx/δa = -x^2 / (2ax + b) So we can calculate also the derivative with respect to b as: ---> δx/δb = -x / (2ax + b) And with respect to c as: ---> δx/δc = -1 / (2ax + b)

@tyronekim3506 2 жыл бұрын

Brilliant! I didn't think beyond the first method. Thanks. I learned something new today.

@sistaseetaram9008 2 жыл бұрын

I watch your videos when I'm bored, your way of solving problems is so enjoyable.

@laelodhner1361 2 жыл бұрын

Fun stuff! You should look up root locus analysis in control theory; this analytic method was invented in 1948 by Walter Evans to do exactly what you're trying to do (track the change in the roots of a polynomial in response to a change in coefficients). Rather than restricting to derivatives, his method lets you actually look at the locus of all roots as some parameter changes.

@christiansmakingmusic777 2 жыл бұрын

Nice. This relationship between x and a has implications for factorization of natural numbers.

@YuriiKostychov 2 жыл бұрын

Hello! And when dx/da=-x^2/(b+2ax)=-(bx+c)/[(b+2ax)*a] here is easer substitute x. or we can even try to solve differential equation :) And naturaly: NO WAR

@weonlygoupfromhere7369 2 жыл бұрын

really cool man. Always making my day with cool math concepts

@blackpenredpen 2 жыл бұрын

Glad to hear. Thanks.

@Nothingx303 10 ай бұрын

Thanks ☺️ for these type of videos

@netsuwan_praphot 2 жыл бұрын

I am going to sit for my country's military acedemy admission test. it is is one of the hardest in my country and this channel really help me a lot solving math quesions😁

@DoodLi-Doo 2 жыл бұрын

1:40-x is a function of a, so it's not a constant. It's the same as saying c=-ax^2-bx, a function of a...

@perspicacity89 10 ай бұрын

Fantastic video, thank you.

@SaurabhKumar-jo6dp 2 жыл бұрын

Sir how to find this integral int -2^3 abs(x)^[x] dx, [.]= Greatest integer function

@SaurabhKumar-jo6dp 2 жыл бұрын

One more integral int sin(x)^cos(x) dx

@calvindang7291 2 жыл бұрын

This first one is pretty easy - split it into 5 integrals and solve them each individually.

@p.g.8796 2 жыл бұрын

Nice discussion! But the statement that "x" is a function of "a" is flawed. The zero crossing is a "function" of a (and also b and c), although I wouldn't use the word function but dependent. And therefore the zero crossing itself is an excplicit term namley 0 = ax^2+bx+c

@richardbloemenkamp8532 2 жыл бұрын

The equation 0 = ax^2+bx+c for given b and c gives a relation between a and x. For certain given domain and range this relation is relation is one-to-one and onto. Then "x" can be given as a function of "a" and "a" can be given as a function of "x". "zero-crossing" more or less suggests that the equation is interpreted as the zero-crossings of a graph of a quadratic. This is a valid interpretation, but there are other interpretations. See Wikipedia "function" for more on functions, implicit functions, multi-valued functions etc.

@pablolinares4895 2 жыл бұрын

I was missing hearing your voice man!! I remember watching your videos for Diff. Eq. They were literally the best!

@DrBarker 2 жыл бұрын

Very fun idea for a problem, I really enjoyed this!

@blackpenredpen 2 жыл бұрын

Thanks!!!

@vitalsbat2310 2 жыл бұрын

wow magical solution!

@flowingafterglow629 2 жыл бұрын

This was cool. Thanks.

@deltadromeus9742 2 жыл бұрын

If you take the limit as a goes to 0 of the quadratic formula, you get the (trivial) formula for solving first degree equations

@minecrafting_il 2 жыл бұрын

Really? That's so cool!

@mathisnotforthefaintofheart 2 жыл бұрын

You can also do that very easily if you apply L'Hospital rule on the quadratic formula (the limit on "a" and consider two cases for positive and negative "b") Only half a sheet of paper...

@curtiswfranks 2 жыл бұрын

I think that ∂(ax²+bx+c)/∂a = x² indeed. In other words, the whole parabola is linear wrt a; as a increases, the parabola widens in direct proportion. What you are doing is seeing how the zeroes of the parabola depend on a, and that is a more complicated question.

@iainpaton6797 2 жыл бұрын

Your original expression was not partial x/partial a, it was partial(ax^2+bx+c)/partial a = partial 0/partial a = 0. The solution to that is x^2=0, so x=0 as you wrote, ie if you have ax^2+bx+c = 0, changing a continues to make the expression equal zero. What you solved is certainly more interesting, but your initial formulation of the problem was correct until you stated it implied partial x/partial a=0.

@karolakkolo123 2 жыл бұрын

Exactly. The way in which the question was formulated, partial derivatives with respect to any of a,b,c, or x are 0, since changing them will not change the result of the equation which is zero by definition

@thomasdalton1508 2 жыл бұрын

Indeed. Talking about differentiating an equation is very sloppy - what you are doing is differentiating both sides of an equation. Sloppy language/notation often leads to mistakes like this.

@rossjennings4755 2 жыл бұрын

This explanation is actually still a bit wrong. If you have ax^2+bx+c = 0, changing a doesn't usually keep the left-hand side equal to zero. In fact, changing a to a+Δa makes the left-hand side equal to x^2 Δa. What solving the equation partial(ax^2+bx+c)/partial a = x^2 = 0 actually tells you is for what value of x it is true that changing a (by a small amount) leaves ax^2+bx+c unchanged. That happens for x=0, because in that case ax^2+bx+c = c, so changing a has no effect.

@poggerschampion2765 2 жыл бұрын

I just love the pokeball, really nails the aesthetic lol

@louisperuso4517 2 жыл бұрын

Love this. Can you please prove that if ax^2+2b'x+c=0, then x=(-b'+/-sqrt((b')^2-ac)/a. This is what I call the EVEN quadratic formula and also can u plz differentiate this if possible?

@calvindang7291 2 жыл бұрын

The proof of this is pretty much trivial; just substitute b=(2b') into the normal quadratic formula.

@guidoreuter6032 Жыл бұрын

Just a GREAT video .. Thank you!!

@user-dc9wf1lv5o 2 жыл бұрын

It seems to me, the first line contains an incorrect statement. There should be not dx/da, but dF/da, where the F = ax^2+bx+c. And if we dont know that the x is a root of the equation (independent parameter) then it will turn out to be true. And if we know that the x is the root of equation, that we get the derivative of an implicit function.

@haihuqin8861 2 жыл бұрын

Yes. Cannot agree more.

@johncalculusmathsclass5998 Жыл бұрын

So brilliant bro

@yanlonghao5012 2 жыл бұрын

The derivitive depends on your setting. If you fix b, c. You get a curve in this vedio. If you fix x, you get a plane, and it was a different story. Without telling anything, it is dangerous to start work just like that.

@Bjowolf2 10 ай бұрын

I was about to suggest that you expressed ðx/ða in terms of x, and then you did - cool 👍 Something interesting must be happening, when D = b^2 - 4ac goes towards zero, as a changes ( for given values of b & c ), and the two roots become identical ( a double root ) - and they then vanish or become complex rather, as D < 0 - a socalled "catastrophe" (Thom) And similar for b & c of course 😉

@reubenmanzo2054 2 жыл бұрын

Your derivative of the quadratic formula can still be simplified. There's a square-root on the denominator which can be rationalised (not mandatory, but a preference for most teachers) and 2 can be factored in the numerator, which cancels a 2 in the denominator.

@GradientAscent_ 9 ай бұрын

It feels good to come back to bprp after a while and watch this video

@ulisespachecosanchez5058 9 ай бұрын

presenting itself in the derivative given in the general equation of the general formula, if you see a difference to what the integration would be, you just see that there are different processes that in themselves must be analyzed more carefully, thank you teacher, the topics are very interesting and aside new for me

@eks_0 2 жыл бұрын

Can you take integral of it next?

@blackpenredpen 2 жыл бұрын

Sure thing!

@Ssilki_V_Profile 2 жыл бұрын

Differentiation of function: F(x, a, b, c)=ax^2+bx+c with respect to a is x^2. So, you plane to differentiate not a polinom function, but it roots.

@sulisueeeeee0553 2 жыл бұрын

Watching this from UK, nostalgia of gcse and Alevel Maths😅😞, But luv the work. After u understand maths, the satisfaction of getting ur answers right when u look through the back of the book is worth it!

@e.s.r5809 2 жыл бұрын

I saw the thumbnail and went "oh this sounds like fun. I wonder what happens if you integrate it though". So I integrated it. Then I sat there for a few moments looking at what I'd wrought. Wondering why I'd done that. What did I gain? What is it even trying to tell me? I should be doing my comms and signals homework, and yet here I am, integrating the quadratic formula. Why.

@blackpenredpen 2 жыл бұрын

😆 I integrated it too So u r not alone.

@jwxq3 2 жыл бұрын

I haven't taken a legitimate calculus class in over a decade, since I was an AP Calculus and Physics student in high school. Now, after covid affecting my industry, my plan is to return to university for an engineering degree. Thank you for uploading this video, very useful and informative. If you don't enjoy calculus, you probably hate sudoku and minesweeper lol.

@kagayakuangel5828 2 жыл бұрын

Beautiful. Cool!!

@ahcenecanpos9463 2 жыл бұрын

very nice calcul

@justsaadunoyeah1234 Жыл бұрын

The Best episode of blackpenredpenbluepen!!!

@SaurabhKumar-jo6dp 2 жыл бұрын

Sir how to find nth derivative of x^x

@blackpenredpen 2 жыл бұрын

You differentiate it n times and look for a pattern.

@SaurabhKumar-jo6dp 2 жыл бұрын

@@blackpenredpen thanks sir

@mathevengers1131 2 жыл бұрын

@@blackpenredpen I tried that but there's no pattern visible to me. I literally spent 8 hours on it.

@SaurabhKumar-jo6dp 2 жыл бұрын

@@mathevengers1131 nice bro by the way I have got the solution.

@mathevengers1131 2 жыл бұрын

@@SaurabhKumar-jo6dp can you please send me the solution

@CK113th121 2 жыл бұрын

This looks like trolling at first, but it gets more and more interesting as the answer starts to form.

@Leberteich 10 ай бұрын

The Quad eqn can always be divided by a (a is never zero otherwise the eqn wouldn't be quadratic). In other words, all a's are 1. It's the b's and c's that may be different from 1.

@jtvanilla1776 10 ай бұрын

All of us at some point have thought of doing many different things to the quadratic equation

@adityasingh4123 2 жыл бұрын

Sir plz make video on full higher calculus

@dr.rahulgupta7573 Жыл бұрын

Sir (d/dx) ( a x^2+bx +c ) = 0 gives x co - ordinate of the parabola y = ax^2+bx +c . (When x is treated as variable and a, b , c as constants .)

@stlemur 2 жыл бұрын

Makes me wonder what the metric for the space whose basis is unit vectors along a, b and c looks like

@rossjennings4755 2 жыл бұрын

This space is the same as the space of quadratic polynomials in x, so a metric doesn't make sense unless we have a way to define distance between quadratic polynomials. But luckily there's a pretty natural choice, which is the root-mean-square distance on some specific interval [x0, x1]: we can define the distance between two polynomials p and q to be ∫ (p(x) - q(x))^2 dx from x=x0 to x1. Given that, you could work out what the metric is. It's probably pretty complicated-looking, but I can tell you without computing it that it will turn out to be the metric of flat 3-dimensional space in disguise, because you can get an orthonormal basis for this space from appropriately scaled and shifted Legendre polynomials.

@salad3806 9 ай бұрын

That was cool

@toaj868 2 жыл бұрын

Since the vertex of a parabola is linearly dependent on b, I suspect the zeroes are too

@theostene4444 2 жыл бұрын

Not a very clean looking answer. Anyways, wanted to say that you helped me gain more interest for math. It has led to me discussing math with my teacher on topics that are harder than what we're learning in class. I belive that is something that pushed my final grade up to A even tough I always got B+ on tests

@nory2441 2 жыл бұрын

Thats why i love math

@aayushjawalekar4169 8 ай бұрын

I love how he casually does it while using a pokeball to speak

@nikitakipriyanov7260 2 жыл бұрын

I differentiated (-b±sqrt(b²-4ac))/(2a) directly. It makes obviously the same answer.

@gingeral253 10 ай бұрын

Melting my brain

@ecohangout4481 2 жыл бұрын

Sir, what would you recommend to learn all high school math? There is no school in my area.

@blakelee4555 2 жыл бұрын

Khan Academy.. it's free and does all math from 1st grade through to vector calculus.. you can pick whatever you want. High school math is just algebra, trigonometry, geometry, and pre-calculus in the United States. Statistics and Calculus are offered as options as well.

@RisetotheEquation 2 жыл бұрын

Next up: my first time differentiating the cubic formula

@binomial9034 2 жыл бұрын

I think that's the first time i see him using the partial derivative simbol xD

@ishanmozur1776 Жыл бұрын

@blackpenredpen mr blackpen red pen i reallly like your sereies especially the five polynomials equation. So please i have a request the request is can you find or give a video of the actual quartic formula which is : ax^4+bx^3+cx^2+dx+e=0 i really NEED that formula for my studies because i study in grade 8 now.

@hoteny 10 ай бұрын

We can write b and c as something something a and x too, just like x was something something b and a. Right? So, we end up in infinitely differentiating them? Because they all end up converting each other.

@iamnotadinosaur9915 2 жыл бұрын

Is there a place I can get that derivatives poster/plaque behind you? I absolutely love it!

@blackpenredpen 2 жыл бұрын

Yes. It’s in my merch store. Link in description. Thanks.

@iamnotadinosaur9915 2 жыл бұрын

@@blackpenredpen Thanks! Will be ordering shortly!

@iamnotadinosaur9915 2 жыл бұрын

@@blackpenredpen I ordered the Calc 2 integrals plaque; super cool! Keep it up!

@danuttall 2 ай бұрын

5:15 The fraction for dx/da should be reduced by dividing each term, top and bottom, by 2.

@jackson9143678 2 жыл бұрын

Can you do this for the cubic formula as well? :)

@thespacetime 2 жыл бұрын

From Master Shifu to a Shaolin Monk. Differentiate that nerds.:D

@dhvalab 2 жыл бұрын

Can you tell me who is given this equation.

@edwardhuang5885 2 жыл бұрын

1:19 "just to keep our sanity," 😂

@kingofdomino0653 9 ай бұрын

When you did the first equation, why didn’t you factor out a two and cancel with the bottom? Wouldn’t that have simplified the first equation even more?

@kushagrasahgal 10 ай бұрын

how can you differentiate w.r.t a as a is constant

@SafetyBoater 2 жыл бұрын

I would like to see the graph of a=1. x-axis=b, y-axis=c, and z-axis is each of the partials. Since all quadratics can be rewritten in the form x^2+(b/a)x+(c/a)=0, I think this should show all rates of change.

@Goten40373 2 жыл бұрын

at 5:20 i think you could factor out a 2 on the bottom and top?

@waltuhputyourdaway6561 2 жыл бұрын

More calc3 content would be nice

@rubensramos6458 Жыл бұрын

To find an analytical solution for ax^2+bx+c = 0 is easy. However, what is the analytical solution for ax^(2+e)+bx+c=0 with ‘e’ being a real number? The solutions are x1=(b/(az))Wq(((-c/b)^z)(a/b)z)^(1/z), where z = (1+e) and q = 1-1/z. x2 = (-y(a/b)Wq((-1/y)(b/a)((-c/a)^(-1/y))))^(-1/(1+e)) where y = (2+e)/(1+e) and q = 1+y Wq is the Lambert-Tsallis function (a generalization of the Lambert function). Sometimes the correct solution is x1, in other cases the correct one is x2 and there are cases where x1 = x2, depending on the values of a, b and c. For example the solution of x^(2.5)+x-1 = 0 is x1 = x2 = 0.6540 (up to 4 decimals).

@engralmightyformula1951 2 жыл бұрын

Wow!!!.😍 This is very surprising.

@mangalesh7936 Ай бұрын

Give this dude a bigger board

@edal7066 2 жыл бұрын

so, in other words you evaluated the sensitivity of the roots on changes in a parameter.

@PinkPastelShark 2 жыл бұрын

Why is x a function of a but b and c are not functions of a? You could solve for either b or c and it would be in terms of a

@blackpenredpen 2 жыл бұрын

It could. Then we would be doing db/da dx/da and all that. Omg… we have a lot of combinations dx/dc, da/dx etc 😆

@Neiltwiss 2 жыл бұрын

I think because x Is a variable While b and c are not. So b and c can take a value that doesn't depend on the value of a, but the opposite for x

@pieTone 2 жыл бұрын

You got sponsored, good job!

@Sg190th 2 жыл бұрын

That's why implicit differentiation is a savior. Scraping that much work off.

@DavidHinkes 2 жыл бұрын

Great video! Question: you assume db/da and dc/da are zero as well. Why is this any different than dx/da, which is not zero? For that matter, what is the difference between a constant and a variable really? I've always found this confusing.

@jeremydavis3631 2 жыл бұрын

In the context of calculus, my understanding is that a constant is any variable whose derivative (with respect to anything) is identically equal to 0. (We don't usually call them variables, but that's the only difference.) As for db/da and dc/da, I think you have a good point. He implicitly assumed that a, b, and c are independent, but that didn't need to be the case. It makes sense when a, b, and c are parameters (variables that are part of the function definition and which calculus treats as constants). But since this exercise blurs the distinction between parameters and dependent variables, the assumption should at least be explicitly stated.

@sk8erJG95 2 жыл бұрын

Those aren't assumptions, the variables a,b,c are called "free variables" because we are free to choose them to be anything. But x is a "dependent variable" because it depends on a,b,c. Something is a "constant" if it doesn't change when any other variables change, i.e. it's a variable that is fixed.

@DavidHinkes 2 жыл бұрын

@@sk8erJG95 Isn't a, b, or c just a function of x as well?

@sk8erJG95 2 жыл бұрын

@@DavidHinkes Absolutely! So in the context of the video, looking at ax^2 + bx + c = 0, you'd normally think about a,b,c as free variables (constants). BUT the quadratic formula tells us there is a relation between x, a, b, and c. Therefore dx/da has to take this into account. For any two variables A, B, you could consider "these variables are independent" to be the same as "dA/dB = 0". It is wierd to think about - that's why he made the video, I think!

@squonkusmcfreengle1584 2 жыл бұрын

yeah it just depends on what you consider fixed and what you consider a variable a way to think about what he is finding in this video: if you vary a in a quadratic equation, the places that the parabola crosses will move according to the equation he found, ie you will need a different value for x to make that equation zero if instead you treat x as a constant independent of a, but c as your variable, it is totally valid to do what he said was wrong in the beginning: dx/da will be x^2, but c will change when you vary a to keep the equation equal to zero. a weird example of this: take the derivative of (ax^2+bx+c)=0 with respect to x repeatedly and you get 2a=0 -> b = 0 -> c=0. this would seem to imply a contradiction if a were nonzero, but it actually doesn’t. by taking the derivatives and assuming equality held in doing so, you assume the equation holds for ANY x- which is only true if all the constants are zero.

@nimmira 2 жыл бұрын

Out of curiosity, what would the rate change of X in terms of its coefficient signify? Like, for example, let's say in Physics?

@Grak70 2 жыл бұрын

One example that comes to mind: given a fixed height above the ground, how quickly does the time it takes to reach that height vary with respect to changing gravitational constant?

@nimmira 2 жыл бұрын

@@Grak70 Thanks! That's very interesting.

@gabrielaudi100 2 жыл бұрын

Accelerations

@Narteur04 2 жыл бұрын

good video, watching you from my car un france haha

@otavioluz2809 2 жыл бұрын

Fantastic

@Latronibus 2 ай бұрын

d(ax^2+bx+c)=x^2 da + x db + dc + (2ax + b) dx, so if that is zero and db and dc are zero then dx/da=-(x^2)/(2ax+b). Plug in x matching the quadratic formula and you're done. It's kind of a nice way to go.

@shahjahonsaidmurodov Жыл бұрын

Note: you can simplify further by halfing the entire fraction

@krrishmaheshwari4860 2 жыл бұрын

What is conclusion? Like why we did it?

@eigenfeynman9890 2 жыл бұрын

The most impressive part about this is, the derivative of this demon is done without any mistake and without rubbing anything.

@gurindersinghkiom1 2 жыл бұрын

Cool 😎 👏

@agrocassiano Жыл бұрын

Boa tarde; X = { -b/2 + - ^[ (-b/2)^2 - (c.a) ] } : a Obs; quando (b) for n.o ímpar, multiplique a equação por .(2), evitará frações... x^2 - 5x + 6 = 0 . (2) 2x - 10x + 12 = 0 X = { 5 + - ^[ 25 - 24 ] } : 2 X' = { 5 + 1 } : 2 X' = 3 X" = { 5 - 1 } : 2 X" = 2 Parabéns Mr Black, sucesso sempre.

@roysmiles1 2 жыл бұрын

But,in the standard quadratic formula, "a" is constant, so da=0 and dx/da is undefined, isn't it? What does it mean to differentiate wrt a constant?

@somerandomashellperson7130 10 ай бұрын

We aren't considering a to be a constant in this scenario, it is a variable (like x usually is). The top of a partial derivative will be a function that contains the variable in the bottom of the partial derivative (at least with my understanding of calc, which is admittedly quite limited). dy (with y being a function with the variable) --- dx (the variable is x)