Proof of the Second Derivatives Test

[@jung-akim9157]

I've been looking for a concrete proof of second derivatives test for the past three days. This is it! Thank you so much! you're so brilliant!

[@rohitmaurya5392]

Finally found the rigorous proof of Second Partial Derivative test. Thank you very much.

[@JPK314]

Thanks for making this video! One comment I would make is that your "h"s look exactly like your "k"s. Also, to motivate the t1=h1/k1 statement, one can look at how g(t) is related to Du(Du(f)): Dividing Du(Du(f)) by k^2 gives us a constant term (which is fyy) and leaves us with h^2/k^2*fxx+2h/k*fxy+fyy. Setting Du(Du(f))/k^2=g(t), we see that h^2/k^2*fxx+2h/k*fxy = t^2*fxx+2t*fxy. By symmetry, this means h/k = t. Because we know (for D < 0) that there exist g(t) > 0 and g(t) < 0, we can make this back into the form Du(Du(f)) by reversing the process: k^2*g(h/k) > 0, and k^2*g(h/k) = Du(Du(f)), so Du(Du(f)) > 0; k^2*g(h/k) < 0 , and k^2*g(h/k) = Du(Du(f)), so Du(Du(f)) < 0. Because we know these t exist, we know the h and k exist as well, which means there indeed are two vectors by which the surface is concave up and concave down respectively.

[@troelsvejenchristensen8850]

Thank you :)

[@mohithalder3169]

Ok.... Found a new channel to binge watch

[@utuberaj60]

Very simple and direct proof. But, hold a sec please. You have used the dot product of the unit vectors and gotten the expression of the 2nd derivative as a quadratic expression (at 5.53 mins) showing a term 2f(xy)*hk. But ma'am, this term should reduce to zero as "h dot k" will be zero. How come you pulled this off?

[@lindagreen7859]

h and k are the components of a unit vector , so when I write hk I just mean their ordinary product as numbers, not a dot product.

[@wunboonail]

Thank you very much, Linda. You have packed hours of reading into 15 mins.

[@strengthinnumberstutoring61]

What a great proof! I do have one question, probably a function of my own incompetence. Why is it sufficient to consider the directional derivative twice in the same direction? Shouldn’t we be requiring it along two linearly independent vectors? I’m sorry if this is an obvious answer, but the relationship between D_{uu} and the “one direction of increase/decrease” behavior isn’t clear to me.

[@lindagreen7859]

I am trying to figure out in what directions the surface is concave down (and has a local max in that direction) or concave up (and has a local min in that direction. Taking the derivative of the derivative in the same direction u is the analog to taking the second derivative for a function of one variable and checks for concavity. In one of the cases, I do consider the derivative of the derivative in two linearly independent directions, to verify that the surface is concave up in one direction and concave down in another.

[@strengthinnumberstutoring61]

Actually I think I made an error in the assumptions that underlie my question. What I was saying was more along the lines of taking the directional derivative in one direction, and then taking the directional derivative in another linearly independent direction. What I still don’t understand is this. It seems to me that after verifying that f_x and f_y are zero, it should suffice to demonstrate that f_xx and f_yy both have the same sign (if positive then local min, if negative then local max). The reasoning is more geometric than analytical: showing that a level curve along constant x has a minimum AND that a level curve along constant y has a minimum should demonstrate that the function has a minimum. (If they fail to have the same sign, then one would have a min and one would have a max - classic saddle point.) I know that this is a “standard” result from multivariable calc, and so I don’t think I’ve cracked a new egg here. But is there an easily-understood reason why it’s not sufficient for both of the repeated second derivatives to be the same sign?

[@lindagreen7859]

@@strengthinnumberstutoring61 Right, at first glance it seems like if f_xx and f_yy are both positive, for example, then you'd have to have concave up behavior everywhere and get a local min. But in fact, there might be a direction that is concave down in between the x-direction and the y-direction. For example, if you graph f(x, y) = x^2+y^2 + 4xy, you'll notice it looks like a saddle at (0,0), even though f_xx = 2 and f_yy = 2 both have the same sign. You'll notice also that f_xx*f_yy - f_xy^2 = 2*2 - (4)^2 is negative ... so the 2nd derivatives test is not fooled!

[@godswillonwugbenu9978]

Please is this for a Laplace transform??

[@martinperu6207]

Please some books for calculus 3 and Vector Calculus...

[@arindamhazarika1905]

Thank you madam. Now I finally understand the second derivative test. Thank you so much

[@liam.config]

why do the k's and h's look identical

[@andeslam7370]

Thanks a lot for you explanation. It's so much better than my professor's.

[@efeguleroglu]

g(t) is negative for some u and positive for some u. But why does it mean that the point is a saddle?

[@JuanVazquez-si9xq]

the double directional derivative is equal to g(t)*(k^2) when t=h/k. So if g(t) can be both positive and negative so can the double directional derivative, which means it can be both "concave up" and "concave down" depending on the direction

[@wishcow]

It is not clear from the proof where the continuity of the second partial derivatives is used.

[@gaudiowen]

It is necessary to use Clairaut-Schwarz theorem about mixed derivatives.

[@Jawu523]

I don’t understand why t1=h/k

[@oyaoya2468]

excellent video. Thanks a lot, you help me clear everything up

[@sammao8478]

flawless piece of work, thank you!

[@ugursoydan8187]

THANK YOU VERY MUCH!!! WHAT A GREAT PROOF!!!!

[@hemnathl]

Thank you very much for the clear explanations.

[@denstupakov]

Brilliant!

[@serkangoktas8753]

thank you for thorough explanation :)

[@blablablerg]

great explanation

[@SoumilSahu]

THANK YOU so much!

[@withoutword2888]

pretty excellent. really helpful. ty

[@gurleenkaur9601]

Thanks ma'am

[@feraudyh]

Good as always.

[@Jawu523]

Why some root is positive some are negative

[@Jawu523]

11:34

[@lokeshverma5902]

wow! i have been getting frustrated learning the proof given in thomas calculus .. you circumvented the most of the theoretical mess as given in the book and made it more intuitive. thanks but you didnt explain the case when discriminant =0

[@lokeshverma5902]

i mean why is that inconclusive

[@jasonli1060]

YAYYY

[@shawzhang4498]

finally