Second partial derivative test intuition

[@yasseralg3928]

This guy is a gift to humanity

[@rajinish0]

It's also sort of helpful to plot a vector field of the gradient function. If d^2f/dx^2 and d^2f/dy^2 are both positive, we'd have divergence around the center that'd make the point a minimum, however, if you plot some vector fields you can see that for saddle points the vectors sort of fly off in top right and bottom left corners leaving the diagonal as a saddle point(if you plot the example in the previous video). That measure of how much they fly off is df/dydx; you could imagine moving in x direction and looking at the change in y component of the vectors, or move along y axis it doesn't really matter. So there's a certain threshold *(d^2f/dx^2)*(d^2f/dy^2) = the mixed partial derivative^2* after which we get a saddle point.

[@justinward3679]

It should be illegal to make calculus this simple.

[@harisahan2126]

Justin Ward absolutely right

[@paul_tee]

good visuals for intuition and lack of proofs go a long way

[@zes7215]

Hari Sahan not, and no such thing as simple or not simple, any can be ok

[@mihaililiev5932]

Quite amazing, innit?

[@pasqualegiusto8306]

hnnn

[@savonprice9821]

Dude??? Why is Khan Academy so freaking amazing!!!!!!

[@Jonisrock00]

I'll never look at pringles in the same way again..

[@pierre7770]

Genius 😂

[@ankiteleciitd]

Another way to think about the Second partial derivative test is using Linear Algebra: Consider a 2x2 Matrix H with elements as [[fxx, fxy], [fxy, fyy]], with derivatives evaluated at (x0,y0) Now, think about how the point (x0, y0) will move once it is exposed to the matrix H. This is a Linear transformation case. If the linear transformation stretches the area onto the same plane, then that point (x0,y0) represents either a maxima or minima. On the other hand, if the linear transformation flips the plane ==> (x0,y0) is a saddle point. Mathematically, if det(H) > 0 ==> maxima/ minima det(H) saddle point det(H) = 0 ==> cant decide.

[@valerianmp]

Ankit Gupta never notices that, thanks!

[@nikhilwat]

The Hessian matrix :)

[@robmarks6800]

Cool, but I do not understand how the flipping of space is equivalent to a saddle point

[@lordcasper3357]

@@robmarks6800 he's just talking because the formula looks like determinant of a 2x2 matrice, nothing related between these 2 topics

[@poiuwnwang7109]

@@robmarks6800 same question here

[@sercantor1]

how can you be this good?

[@thelastcipher9135]

This should be called the Pringle Test.

[@Masteridea101]

i owe my whole tuition fee for your videos man... U guys are rocking.. God bless you

[@fahadashrafofficial]

I wish my teachers were more like you :)

[@00rq34]

Can you please provide a link to your paper?

[@TheHuggableEmpire]

I thought I was too dumb for any form of teaching and got my mind ready for not understanding anything. But your video headshotted all of my doubts to my pleasant surprise. Layman term yet fundamentally solid explanations like yours are what I need

[@huan320]

omg!! the series of multivariable function is incredibly clear!!! OH my god!! YOU are a talented teacher when doing a job like this!

[@NicolasSchmidMusic]

Where can we find that article you are talking about? It would be nice to put it in the description

[@هيلة-ع8م]

where can i fine the article that he mentioned?

[@Nerdwithoutglasses]

As Grant said, there is a more rigorous way to show why the test has that form. From the formula of total derivative df, we can construct a formula for d^2(f) which is a quadratic form of dx and dy. Then, we can classify the critical point based on the sign of d^2(f) (it is a saddle point if d^2(f) can be positive or negative; it is a maximum or a minimum if d^2(f) is always positive or always negative). And, determining the sign posibilities of d^2(f) by the discriminant of the quadratic form- the minus H.

[@sarthakmaheshwari3888]

Hats off to your teaching skills

[@poiuwnwang7109]

@8:10, the height alone x axis is a constant, which corresponds to df/dx=0, not d2f/dx2=0, like the tutor said. Anyboday any comments?

[@mridulk81]

df/dx = 0 also implies that d²f/dx² is also 0

[@ralphhebgen7067]

Very cool. Actually, if there is disagreement between fxx and fyy, that is sufficient to conclude it is a saddle point. Only if there is disagreement do we need to deduct fxy to see whether the gradients that are not strictly along the x- or y-axis identify the point as a saddle.

[@anilphiyak9862]

Can u tell me the app u use fro graph in 3d

[@berkeunal5773]

Why do we only consider xy term? There are several terms (such as xy^2) which have non-zero mixed partial derivatve. Why this theorem holds for them too?

[@mridulk81]

anybody got the link to the article mentioned at 9:42?

[@BD-dh9fw]

Also worth pointing out that H is just the determinant of the Hessian matrix, which is, I'm assuming, why he labeled that variable as H. Kind of surprised he didn't identify that explicitly in the video, because it makes it easier to remember the test, knowing that we're just looking at the sign of the determinant of the Hessian. It also enforces the parallel with the single-variable case, where we're looking at the sign of the second derivative, because we can imagine the Hessian as analogous to a multivariable second derivative.

[@khalilmohammed2297]

could you tell us please where is the Article you have mentioned .

[@MagnificentCreature]

Amazing video that clarifies a big ? I had when learning this

[@yadinandyanay]

Can u just say it’s the determinant of the hessian for future reference? Wish I had known sooner

[@tototoday7609]

xy 성분이 커질수록, 대각선 방향에서 포물선을 만드는 경향이 커진다는 건가..?

[@blink1747]

Yes, critical points exist even if they are not any of extrema both in 2D & 3D.

[@swarajyalaxmi8692]

where can i find the article

[@charmendro]

Bruh i wish i saw these while i took multivariabel calc

[@alvarol.martinez5230]

Where are the articles you mention?

[@luffyorama]

I think the articles are in Khan Academy site.

[@alvarol.martinez5230]

Oh thanks!

[@anonxnor]

I think it's this www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/second-partial-derivative-test

[@rodrigo_t9]

Here's directly the actual full reasoning article, which is linked on that one: www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/optimizing-multivariable-functions/a/reasoning-behind-the-second-partial-derivative-test

[@barnikroy9413]

@@rodrigo_t9 thanks a lot...!

[@lordcasper3357]

can't find the related link about the formula, is it removed?

[@kavitakohli5632]

Why we always check the sign of partial double derivative of x only and not y to determine max or min when that h is positive

[@shayanmoosavi9139]

Because we know that if H is positive it's either a max or a min. The (fxy)^2 term is always positive so we're always subtracting a positive number (adding a negative number if you like) and the only way for H to be positive in this condition is if the first term is positive so either both of them (fxx and fyy) are positive or both of them are negative therefore if we know the sign for either of them it'll be enough. Hope that helped.

[@liffidmonky1216]

Where is the article??? :( i want it xd

[@mridulk81]

did you find it?

[@harry_page]

I'm surprised Grant didn't say something like "the mixed second partial derivative is a kind of diagonal saddlification of the surface" :D

[@Rockyzach88]

What if your 2nd partial derivatives equal 0? Not H, but fxx. How can you absolutely have a max/min but not have concavity? What does that mean? Or did I just make a mistake?

[@GonzieGr01]

Rockyzach88 if fxx and/or fyy are equal to cero, then the first term of H is going to be cero, making H less than cero (assuming fxy exists) and, therefore, indicating that you are in a saddle point

[@carultch]

You can only conclude you have a saddle point if H < 0. If H=0, the 2nd derivative test is inconclusive. If both direct 2nd partial derivatives equal zero, then this means you simply don't have curvature in the x and y directions, and all your curvature is on a direction that is rotated from them. The mixed derivative would tell you the information about the function's curvature. If both fxx and fyy are equal to zero, but fxy and fyx are not, then you have a saddle point, where your most-positive and most-negative curvatures will occur in rotated axes from the x and y directions. You can construct an equivalent of Mohr's circle, to find those directions, and find the principal second partial derivatives where curvature is maximum. If the entire 2nd derivative matrix has a determinant of zero, then this means you have a situation where the 2nd partial derivatives aren't good enough to tell you the nature of the stationary point. You could have an inflection point, or you could have the nature of the stationary point buried in a higher order derivative, like it is in the single variable case of y=x^4.

[@eamonnsiocain6454]

Grant Sanderson of 3Blue1Brown

[@karthiksukumaran85]

You are just amazing man!

[@gemacabero6482]

Why is the 2nd partial derivative with respect to x twice the one that determines if it is a maximum or a minimum ?

[@SamiDoustdar]

It’s not, doing it with y would work too. It’s just if one is positive so would the other be because this is in the scenario where the point is either a minima or a maxima meaning both the x and y planes have the same concavity so only testing one is enough

[@tantarudragos]

is this the 3blue1brown guy(cant remember his name)? their voices sound sooo similar

[@aidansgarlato9347]

Yes there the same person

[@SohamChakraborty42069]

Yes, and his name is Grant Sanderson.

[@rameezwaniii]

but why double?

[@sefalipanda3448]

Which playlist is this in???? @khanacademy

[@ziadyoussef9157]

i think multivariable calculus

[@learningsuper6785]

Why are you assuming fxy at x0, y0 is the same as fyx at x0, y0? Isn't that something that requires explanation because it is not obvious.

[@gonk9336]

Look up Clairaut's theorem on mixed partials

[@wongbob4813]

take a look at video 18 in the series: kzbin.info/www/bejne/gGGbXn9ol9qcna8

[@marcustrevor1883]

He talks about mixed partials in another video

[@eswyatt]

Makes absolutely no mention that the second derivative test is the determinant of the Hessian matrix, which he introduced 7 videos ago in talking about the much less important quadratic approximation!

[@anthonyymm511]

Yea this stuff makes a lot more sense if you know that the determinant of a matrix is the product of eigenvalues and the trace is the sum of eigenvalues.

[@pierre7770]

Bro i love you

[@lkc9599]

Why fxy not fyx

[@isakwatz11]

they are the same so long as the second partial derivative is continuous

[@bite027_ketansharma6]

i LOVE you.

[@lauragrandas4036]

I love you.