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## Multivariable calculus

### Course: Multivariable calculus > Unit 3

Lesson 3: Optimizing multivariable functions- Multivariable maxima and minima
- Find critical points of multivariable functions
- Saddle points
- Visual zero gradient
- Warm up to the second partial derivative test
- Second partial derivative test
- Second partial derivative test intuition
- Second partial derivative test example, part 1
- Second partial derivative test example, part 2
- Classifying critical points

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# Second partial derivative test intuition

The second partial derivative test is based on a formula which seems to come out of nowhere. Here, you can see a little more intuition for why it looks the way it does. Created by Grant Sanderson.

## Want to join the conversation?

- where is the written form of that article you talked about?(13 votes)
- (13 votes)

- I know this video is meant with the best intention but I feel the intuition for the dxdy (cross term) was not explained well. Can someone clarify it to me why it matters and why it measures how "similar" we are to f(x,y) = xy?(4 votes)
- I think the intuition is that if we check concavity along only the x-input and y-input, we may get what appears to be a consistent result. For example, they may both have second partial derivatives that are positive, indicating the output is concave up along both axes. However, if we look at the concavity along inputs that include both x and y (ie dxdy), it could be revealed that the concavity is not consistent and we may have a saddle point. I'm not sure what you mean about "how 'similar' we are to f(x,y) = xy."(7 votes)

- (Might be helpful for future readers) the point is just that for the point to be a maxima or minima, the directional second derivative always has the same sign, u.Del u.Del f = 0 has no roots for unit vectors u. You can rewrite this as u^T H u = 0 having no roots.

If you simplify this, setting the y-component of u, u_y as sqrt(1-u_x^2) you get a quadratic in u_x. So you look at the discriminant of this, which simplifies to f_xy^2*(f_xy^2 - f_xx f_yy).

If this is positive (i.e. the term given in the video is negative), it means you have two roots, which give you the points where the double-derivative flips sign, and the point is a saddle point.

If it is negative, it has no roots, and therefore the sign is always the same, i.e. all the second directional derivatives agree.

If the discriminant is zero, it means that there is exactly one direction in which the double-derivative is zero, i.e. you have a straight line (since the directional derivative is also zero, this means the straight line is flat, parallel to the xy-plane like in the video).(5 votes) - It seems to me that the f_xy^2 term is really just (f_xy)(f_yx) in disguise. The fact that f_xy = f_yx is a convenient computational coincidence, but it seems conceptually inconsistent with the (f_xx)(f_yy) term.(3 votes)
- What would really help Grant's graphs, and make things clearer, is if he he had big clear, X,Y, and Z labels on his axes, and even if he color coded the axes and their labels on top.(3 votes)
- Will it be a saddle point if the second partial derivative of x and the second partial derivative of y have different signs? As in one is positive and the other is negative?

If this is the case then we probably won't have to check the mixed partial derivative.(2 votes)- Yes, if you evaluated at H this would always come-out to be negative.(2 votes)

- What if we have more than two variables, how to know either it is max, or min or saddle(2 votes)
- I think you have to calculate the determinant of the Hessian matrix for that equation, which is basically what he's doing here for a two variable function.(2 votes)

- Is there any reason why the expression used in the test is the determinant of the Hessian matrix?(2 votes)
- I'm not sure why the interesting parts here was skipped and why he skipped the important reason for that specific formula, I get that he tries to explain it without using Linear Algebra, which is strange as it was a prerequisite for this course. Anyway, let me give you a brief intuition.

The whole reason why we have that formula is because it is the DETERMINANT of the Hessian matrix. Why you may ask? Because for a diagonalisable matrix, where we have a a linear independent eigenbasis, the determinant of the matrix equals the product of the eigenvalues. So what does this have to do with second partial derivatives and the Hessian?

Well, if you remember the quadratic approximation of a function? That will look very similar to the function at a specific point, in this case a critical point and will have a very similar shape to a maximum or minimum point as parabola shaped functions can perfectly "hug" the graph.

So why is this important? This comes from the quadratic form from Linear Algebra (x^T)Ax where A is a symmetric matrix and guess what? The Hessian is a symmetric matrix. Therefore all the properties of the quadratic form applies in this case.

If we approximate a function using the Taylor series at a critical point, the first derivative will of course be zero so we will have something like T(p) = f(p) + (p - p_0)^T H p. As we are basically only after the "form" of the function (and by form I mean how it looks around that point), we can remove all the shifting up/down & left/right which will place it at the origin and we will end up with the standard quadratic form from Linear Algebra: T(p) =(p)^T Hp.

Without going to deep here as it would require explaining change of basis (another large topic), but there exists a diagonal representation of the Hessian matrix in another basis such as:

(p)^T Hp = (q)^T D q where q is the coordinate vector of p. Anyway, guess what values D contains? The eigenvalues! This means that the eigenvalues affects q directly (after multiplication) and as q will always be positive because we have a diagonal matrix and therefore all q's will be squared, the eigenvalues needs to have the same sign in order to "move" in the same direction i.e the function will have all positive or negative points and be definite.

In order to get a negative point from q, the corresponding eigenvalue needs to be negative and if we have different signs for the eigenvalues, we will have different points at the quadratic function being either positive or negative, i.e a indefinite function or saddle point.

As the determinant is the product of the eigenvalues, the only way to get a negative value is if the signs are different, else we will always get a positive value. We can of course also get zero which is the case when The Hessian is linear dependent and we can't create a basis from the eigenvectors. I guess zero can be viewed as the function is neither concave or convex in both directions, either only in one or in none (flat plane). And btw, the eigenvalues directly affects how concave or convex a quadratic function is.(2 votes) - What if the function does not satisfy the Laplace's equation for the specific point we are looking at (dxdy does not equal to dydx)? What term should we use in the test? I think it might be okay to do dxdx*dydy-dxdy*dydx but then the second term can be negative, I'm confused(2 votes)
- Check out Clairaut's Theorem on the symmetry of second derivatives. Take the partial derivative wrt x then same wrt y: it'll give you the same answer as when you do y first, then x. This theorem also extends to higher partial derivatives.(1 vote)

## Video transcript

- [Voiceover] Hey everyone. So in the last video I introduced this thing called the second
partial derivative test, and if you have some kind
of multivariable function or really just a two variable function is what this applies to, something that's f of x, y and it outputs a number. When you're looking
for places where it has a local maximum or a local
minimum, the first step, as I talked about a few videos ago, is to find where the gradient equals zero and sometimes you'll hear these called critical points or stable points, but inputs where the gradient equals zero and that's really just a way of compactly writing the fact that all the partial derivatives are equal to zero. Now when you find a point like this, in order to test whether
it's a local maximum or a local minimum or a saddle point without actually looking at the graph, 'cause you don't always have the ability to do that at your
disposal, the first step is to compute this long value, and this is the thing I
wanna give intuition behind. Where you take all three
second partial derivatives, the second partial
derivative with respect to x, the second partial
derivative with respect to y and the mixed partial derivative where first you do it with respect to x, then you do it with respect to y. And you compute this
value where you evaluate each one of those at your critical point and you multiply the two pure
second partial derivatives and then subtract off the square of the mixed partial derivative and again, I'll give intuition
for that in a reason, but right now we just kinda take it, oh, alright, I guess
you compute this number and if that value H, if that value H is greater than zero, what it tells you, what it tells you is
that you definitely have either a maximum or a minimum. So you definitely have either
a maximum or a minimum. And then to determine which one you just have to look at the
concavity in one direction. So you'll look at the
second partial derivative with respect to x for example, and if that was positive
that would tell you when you look in the x direction there's a positive
concavity, if it was negative it would mean a negative concavity. And so that means a positive value for that second partial derivative would mean a local minimum
and a negative value would mean a local maximum. So that's what it means if this value H turns out to be greater than zero. And if this value H turns
out to be less than zero, strictly less than zero,
then you definitely have a saddle point, saddle point. Which is neither a maximum, nor a minimum. It's kind of like there's disagreement in different directions over whether it should be a maximum or a minimum. And if H equals zero, the
test isn't good enough. You would have to do something
else to figure it out. So why does this work? Why does this seemingly
random conglomeration of second partial
derivatives give you a test that let's you determine what type of stable point you're looking at? Well let's just understand
each term individually. So this second partial
derivative with respect to x, since you're taking
both partial derivatives with respect to x,
you're basically treating the entire multivariable function as if x is the only variable and
y was just some constant. So it's like you're only looking at movement in the x direction. So in terms of a graph,
let's say we've got like, this graph here, you
can imagine slicing this with a plane that represents movement purely in the x direction, so that'll be a constant y value slice,
and you take a look at the curve where this
slice intersects your graph. And in the one that I have pictured here it looks like it's a positive concavity. So this term right here kind
of tells you x concavity. So it's kind of like the,
what is the concavity as far as the variable x is concerned. And then symmetrically, this over here, when you take the partial derivative with respect to y two times in a row, it's like you're ignoring the fact that x is even a variable
and you're looking purely at what movement in
the y direction looks like. Which on the graph that
I have pictured here, also happens to give you kind of this positive concavity parabola look, but the point is that
the curve on the graph that results from looking at movement purely in the y direction can be analyzed just looking at this partial derivative with respect to y twice in a row. So that term kind of tells
you y concavity, y concavity. Now first of all, notice what happens if these disagree. If say, x thought there
should be positive concavity and y thought there should
be negative concavity. Here, I'll write that
down, what that means. If x thinks there's positive concavity we have here some kind of positive number that I'll just write as like,
a plus sign in parenthesis. And then this here, y concavity, would be some kind of negative number,
so we'll just put like, a negative sign in parenthesis. So that would mean this very first term would be a positive times a negative and that first term would be negative. And now the thing that
we're subtracting off, I'll get to the intuition behind this mixed partial derivative term in a moment, but for now you can notice
that it's something squared, it's something that's
always a positive term. So you're always subtracting
off a positive term which means if this
initial one is negative, the entire term H is
definitely gonna be negative, so it's gonna put you over into
this saddle point territory. Which makes sense,
because if the x direction and the y direction disagree on concavity that should be a saddle point. The quintessential example here is when you have the function f of x, y is equal to x squared minus y squared, x squared minus y squared. And the graph of that, by the way, the graph of that would look like this where, let's see, so
orienting myself here, moving in the x direction
you have kind of, positive concavity which corresponds to the positive coefficient
in front of x squared, and in the y direction it
looks like negative concavity. Corresponding to that negative coefficient in front of the y squared. So when there's disagreement among these, the test ensures that we're
gonna have a saddle point. Now what about if they agree, right, what if either it's the case that x thinks there should be positive concavity and y thinks there should
be positive concavity, or they both agree that there should be, you know, negative concavity. In either one of these cases, when you multiply them
together they're positive. So it's kind of like saying, if you look purely in the x direction or
purely in the y direction, they agree, that there
should be, you know, definitely positive concavity or definitely negative concavity. So that entire first term
is going to be positive. So it's kind of like a clever way of capturing whether
or not the x directions and y directions agree. However, the reason that it's not enough is 'cause in either case we're still subtracting off something
that's always a positive term. So when you have this agreement between the x dicretion
and the y direction it then turns into a battle between this x, y agreement
and whatever's going on with this mixed partial derivative term. And the stronger that mixed
partial derivative term, the bigger this negative number, so the more it's pulling the entire value H towards being negative. So let me see if I can give a little bit of reasoning behind why this mixed partial derivative term is trying to pull things towards
being a saddle point. Let's take a look at
the very simple function f of x, y, is equal to x times y. So what that looks like graphically, f of x, y equal x times y, is this. It looks like a saddle point. So let's go ahead and look
at it's partial derivatives. So the first partial derivatives, partial with respect to x and partial with respect to y, well when you do it with respect to x, x
looks like a variable, y looks like a constant,
it's just that constant y. And when you do it with respect to y it goes the other way around. Y looks like the variable,
x looks like the constant so the derivative is that constant x. Now when you take the
second partial derivatives, if you do it with respect
to x twice in a row you're differentiating
this with respect to x, that looks like a
constant, so you get zero. And similarly, if you
do it with respect to y twice in a row, you're doing this and the derivative of x with respect to y, x looks like a constant, goes to zero. But the important term, the one that we're getting an intuition about here, this mixed partial derivative,
first with respect to x then with respect to y, well
you can view it in two ways. Either you take the
derivative of this expression with respect to y, in which case it's one, or you think of taking the
derivative of this expression with respect to x, in
which case it's also one. So it's kind of like this
function is a very pure way to take a look at what this mixed partial derivative term looks like. And the higher the coefficient here, if I had put a coefficient
of, you know, three here that would mean that the
mixed partial derivative would ultimately end up being three. So notice, the reason that
this looks like a saddle isn't because the x and
y directions disagree, in fact if you take a
look at pure movement in the x direction it just
looks like a constant. The height of the graph along this plane, along this line here is just a constant which corresponds to the fact that the second partial derivative
with respect to x is equal to zero. And then likewise, if
you cut it with a plane representing a constant x value, meaning movement purely
in the y direction, the height of the graph doesn't
really change along there, it's constantly zero which corresponds to the fact that this other
partial derivative is zero. The reason that the whole
thing looks like a saddle is 'cause when you cut it
with a diagonal plane here, a diagonal plane, it looks
like it has negative concavity. But if you were to chop it,
you know, in another direction it would look like it
has positive concavity. So in fact, this xy term
is kind of like a way of capturing whether there's disagreement in the diagonal directions. And one thing that might
be surprising at first is that you only need one of these second partial derivatives in order to determine all of the information about the diagonal directions. 'Cause you can imagine,
you know, maybe there's disagreement between movement
along one certain vector and movement along
another and you would have to account for infinitely many directions and look at all of them. And yet evidently, it's
the case that you only really need to take a look at this mixed partial derivative term. You know, along with the original pure second partial derivatives with respect to x twice and with respect to y twice. But still, looking at
only three different terms to take into account possible disagreement in infinitely many directions actually feels like quite the surprise. And if you want the full,
rigorous justification for why this is the case, why this second partial derivative test works and kind of, an airtight argument. I've put that in an
article that you can find that kind of, goes into the dirty details for those who are interested. But if you just want the intuition, I think it's fine to think about the fact that this
mixed partial derivative is telling you how much your function looks like the graph of f
of x, y equal x times y. Which is the graph that kind of captures all of the diagonal disagreement. And then when you let that term, that mixed partial derivative term, kind of compete with the agreement between the x and y directions, you know, if they agree very strongly, you have to subtract
off a very strong amount in order to pull it
back to being negative. So this battle back and
forth, if it's pulled to be very negative that
will give you a saddle point, if it doesn't pull hard
enough, then the agreement between the x and y directions wins out and it's either a local
maximum or a local minimum. So hopefully that sheds
a little bit of light on why this term makes sense
and why it's a reasonable way to combine the three different
second partial derivatives available to you, and
again, if you want the full details, I've written that
up in an article form. I'll see you next video.