We often get the type of problem on our exams where a point (x,y) gives H=0. We are then told to use the "definition" of a saddle point to check if this is the case. My teacher used an example where the point was (0,0)=0, and the function f(x,y) = x^2 y + xy^3 + xy^2. He then replaced (0,0) --> (a,a), which in turn made f(a,a)= a^3(a+2) and chose a just above and below zero. Since f>0 when a>0 and f<0 when a<0, the conclusion was that the point (0,0) was a saddle point. If you've made it this far, I applaud you. Now for my question: If we apply the same test to a max, would BOTH "(a,a)" then gives us values just below the value of f at the actual point? Likewise, would the test give two values just ABOVE the value of f at the actual point, if the point was a minimum? This seems right to me. If the graph looked like a traffic cone, all points below the max would compute smaller values of f, right?

Great question! Can we get an answer on this one?

How do I find the second partial derivatives for a function with 3 variables and how does this test work for that? Thanks :)

You actually need to look at the eigenvalues of the Hessian Matrix, if they are all positive, then there is a local minimum, if they are all negative, there is a local max, and if they are of different signs, then there is a saddle.

What do you do if the second partial derivative still has variables in it? How do you know if fxx or fyy is positive or negative?

If the second partial derivative is dependent on x and y, then it is different for different x and y. fxx(0, 0) is different from fxx(1, 0) which is different from fxx(0, 1) and fxx(1, 1) and so on. There's nothing wrong with that. You need to decide which point you care about and plug in the x and y values. Recall that was also the case with the second derivative test in single var calculus. You calculate the first or second derivative at some point.

What if H>0 but fxx=0?

This isn't possible Since fxy=fyx, (fxy)*(fyx) must be positive or 0. If fxx=0, then (fxx)*(fyy)=0. The determinant H is (fxx)*(fyy)- (fxy)*(fyx), so if fxx=0, then H cannot be greater than 0. Hope this helps :)

We were taught that the test actually was H = (f_12)^2 - (f_11)(f_22) and that if H > 0, the point is a saddle point and of course the opposite for local max/min. This contradicts what you have written here.

It's absolutelly OK. lets say... fxx*fyy - (fxy)^2 > 0 meaning it's max/min. What you have provided here is this inequality*(-1). Proof: -(fxx*fyy) + (fxy)^2 < 0 is still a local maximum/minimum. Was my explanation clear enough?

Main content

Multivariable calculus

Course: Multivariable calculus > Unit 3

Lesson 4: Optimizing multivariable functions (articles)

Second partial derivative test

Q: We often get the type of problem on our exams where a point (x,y) gives H=0. We are then told to use the "definition" of a saddle point to check if this is the case. My teacher used an example where the point was (0,0)=0, and the function f(x,y) = x^2 y + xy^3 + xy^2. He then replaced (0,0) --> (a,a), which in turn made f(a,a)= a^3(a+2) and chose a just above and below zero. Since f>0 when a>0 and f<0 when a<0, the conclusion was that the point (0,0) was a saddle point. If you've made it this far, I applaud you. Now for my question: If we apply the same test to a max, would BOTH "(a,a)" then gives us values just below the value of f at the actual point? Likewise, would the test give two values just ABOVE the value of f at the actual point, if the point was a minimum? This seems right to me. If the graph looked like a traffic cone, all points below the max would compute smaller values of f, right?

Great question! Can we get an answer on this one?

Q: How do I find the second partial derivatives for a function with 3 variables and how does this test work for that? Thanks :)

You actually need to look at the eigenvalues of the Hessian Matrix, if they are all positive, then there is a local minimum, if they are all negative, there is a local max, and if they are of different signs, then there is a saddle.

Q: What do you do if the second partial derivative still has variables in it? How do you know if fxx or fyy is positive or negative?

If the second partial derivative is dependent on x and y, then it is different for different x and y. fxx(0, 0) is different from fxx(1, 0) which is different from fxx(0, 1) and fxx(1, 1) and so on. There's nothing wrong with that. You need to decide which point you care about and plug in the x and y values. Recall that was also the case with the second derivative test in single var calculus. You calculate the first or second derivative at some point.

Google Classroom

Learn how to test whether a function with two inputs has a local maximum or minimum.

Background

Not strictly necessary, but used in one section:

The Hessian matrix

Also, if you are a little rusty on the second derivative test from single-variable calculus, you might want to quickly review it here since it's a good comparison for the second partial derivative test.

[Quick review of second derivative test]

The statement of the second partial derivative test

If you are looking for the local maxima/minima of a two-variable function f, left parenthesis, x, comma, y, right parenthesis, the first step is to find input points left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis where the gradient is the start bold text, 0, end bold text vector.

del, f, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, equals, start bold text, 0, end bold text

These are basically points where the tangent plane on the graph of f is flat.

[See a picture]

The second partial derivative test tells us how to verify whether this stable point is a local maximum, local minimum, or a saddle point. Specifically, you start by computing this quantity:

H, equals, start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612, minus, start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0d923f, squared

Then the second partial derivative test goes as follows:

If H, is less than, 0, then left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis is a saddle point.
[See a picture]

If H, is greater than, 0, then left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis is either a maximum or a minimum point, and you ask one more question:
- If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, is less than, 0, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis is a local maximum point.
  [See a picture]
- If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, is greater than, 0, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis is a local minimum point.
  [See a picture]
(You could also use start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612 instead of start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, it actually doesn't matter)

If H, equals, 0, we do not have enough information to tell.

[Phrased using the Hessian determinant]

Loose intuition

\underbrace{\overbrace{\blueD{f_{xx}(x_0,y_0)}}^{\begin{array}{c}\scriptsize\text{Concavity}\\\scriptsize\text{in x-direction}\end{array}}\overbrace{\redD{f_{yy}(x_0,y_0)}}^{\begin{array}{c}\scriptsize\text{Concavity}\\\scriptsize\text{in y-direction}\end{array}}}_{\begin{array}{c}\scriptsize\text{Positive only when x and y}\\\scriptsize\text{directions agree on concavity direction}\end{array}}-\underbrace{\greenD{f_{xy}(x_0,y_0)^2}}_{\begin{array}{c}\scriptsize\text{How much }f\text{ looks}\\\scriptsize\text{like }g(x,y)=xy\end{array}}

Focus first on this term:

You can think of it as cleverly encoding whether or not the concavity of f's graph is the same in both the x and y directions.

For example, look at the function

f, left parenthesis, x, comma, y, right parenthesis, equals, x, squared, minus, y, squared

Khan Academy video wrapper

See video transcript

This function has a saddle point at left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, 0, comma, 0, right parenthesis. The second partial derivative with respect to x is a positive constant:

\begin{aligned} \blueE{f_{xx}(x, y)} &= \dfrac{\partial}{\partial x}\dfrac{\partial}{\partial x} (x^2 - y^2) \\ \\ &= \dfrac{\partial}{\partial x} 2x \\ \\ &= 2 >0 \end{aligned}

In particular, start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #0c7f99, equals, 2, is greater than, 0, and the fact that this is positive means f, left parenthesis, x, comma, y, right parenthesis looks like it has upward concavity as we travel in the x-direction. On the other hand, the second partial derivative with respect to y is a negative constant:

\begin{aligned} \redE{f_{yy}(x, y)} &= \dfrac{\partial}{\partial y}\dfrac{\partial}{\partial y} (x^2 - y^2) \\\\ &= \dfrac{\partial}{\partial y} -2y \\\\ &= -2 < 0 \end{aligned}

This indicates downward concavity as we travel in the y-direction. This mismatch means we must have a saddle point, and it is encoded as the product of the two second partial derivatives:

Since start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #0d923f, squared can only be positive, subtracted it will only make the full expression more negative.

start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612, minus, start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0d923f, squared

On the other hand, when the signs of start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99 and start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, y, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612 are either both positive or both negative, the x and y directions agree about what the concavity of f should be. In either of these cases, the term start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #bc2612 will be positive.

But this is not enough!

The start color #0d923f, f, start subscript, x, y, end subscript, squared, end color #0d923f term

Consider the function

\begin{aligned} f(x, y) = x^2 + y^2 + \greenE{p}xy \end{aligned}

where start color #0d923f, p, end color #0d923f is some constant.

Concept check: With this definition of f, compute its second derivatives:

start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, comma, y, right parenthesis, end color #0c7f99, equals

start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, x, comma, y, right parenthesis, end color #bc2612, equals

start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, x, comma, y, right parenthesis, end color #0d923f, equals

Because the second derivatives start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #0c7f99 and start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #bc2612 are both positive, the graph will appear concave up as we travel in either the pure x direction or the pure y direction (no matter what start color #0d923f, p, end color #0d923f is).

However, watch the following video where we show how this graph changes as we let the constant start color #0d923f, p, end color #0d923f vary from 1 to 3, then back to 1:

Khan Academy video wrapper

See video transcript

What's going on here? How can the graph have a saddle point even though it is concave up in both the x and y directions? The short answer is that other directions matter too, and in this case, they are captured by the term start color #0d923f, p, end color #0d923f, x, y.

For example, if we isolate this x, y term and look at the graph of g, left parenthesis, x, comma, y, right parenthesis, equals, x, y, here's what it looks like:

It has a saddle point at left parenthesis, 0, comma, 0, right parenthesis. This is not because the x and y directions disagree about concavity, but instead because the concavity appears positive along the diagonal direction

\left[\begin{array}{c} 1 \\ 1 \end{array} \right]

and negative in the direction

\left[\begin{array}{c} -1 \\ 1 \end{array} \right]

Let's see what the second derivative test tells us about the function f, left parenthesis, x, comma, y, right parenthesis, equals, x, squared, plus, y, squared, plus, start color #0d923f, p, end color #0d923f, x, y. Using the values for the second derivatives you were asked to compute above, Here's what we get:

start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #bc2612, minus, start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #0d923f, squared, equals, left parenthesis, start color #0c7f99, 2, end color #0c7f99, right parenthesis, left parenthesis, start color #bc2612, 2, end color #bc2612, right parenthesis, minus, start color #0d923f, p, end color #0d923f, squared

When p, is greater than, 2, this is negative, so f has a saddle point. When p, is less than, 2, it is positive, so f has a local minimum.

You can think of the quantity start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0d923f as measuring how much the function f looks like the graph of g, left parenthesis, x, comma, y, right parenthesis, equals, x, y near the point left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.

Considering how many directions have to agree with each other, it is actually quite surprising that we only need to consider three values, start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #0c7f99, start color #bc2612, f, start subscript, y, y, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #bc2612 and start color #0d923f, f, start subscript, x, y, end subscript, left parenthesis, 0, comma, 0, right parenthesis, end color #0d923f.

The next article gives more detailed reasoning behind the second partial derivative test.

Summary

Once you find a point where the gradient of a multivariable function is the zero vector, meaning the tangent plane of the graph is flat at this point, the second partial derivative test is a way to tell if that point is a local maximum, local minimum, or a saddle point.
The key term of the second partial derivative test is this:

If H, is greater than, 0, the function definitely has a local maximum/minimum at the point left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.
- If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, is greater than, 0, it is a minimum.
- If start color #0c7f99, f, start subscript, x, x, end subscript, left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis, end color #0c7f99, is less than, 0, it is a maximum.
If H, is less than, 0, the function definitely has a saddle point at left parenthesis, x, start subscript, 0, end subscript, comma, y, start subscript, 0, end subscript, right parenthesis.
If H, equals, 0, there is not enough information to tell.

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Johannes
Posted 7 years ago. Direct link to Johannes's post “We often get the type of ...”
We often get the type of problem on our exams where a point (x,y) gives H=0.
We are then told to use the "definition" of a saddle point to check if this is the case. My teacher used an example where the point was (0,0)=0, and the function f(x,y) = x^2
y + xy^3 + xy^2. He then replaced (0,0) --> (a,a), which in turn made f(a,a)= a^3(a+2) and chose a just above and below zero. Since f>0 when a>0 and f<0 when a<0, the conclusion was that the point (0,0) was a saddle point.

If you've made it this far, I applaud you.
Now for my question: If we apply the same test to a max, would BOTH "(a,a)" then gives us values just below the value of f at the actual point?
Likewise, would the test give two values just ABOVE the value of f at the actual point, if the point was a minimum?

This seems right to me. If the graph looked like a traffic cone, all points below the max would compute smaller values of f, right?
Button navigates to signup pageComment on Johannes's post “We often get the type of ...”
(12 votes)
Answer
- Will Simon
  Posted 6 years ago. Direct link to Will Simon's post “Great question! Can we ge...”
  Great question! Can we get an answer on this one?
  Button navigates to signup page
  (3 votes)
Will Simon
Posted 6 years ago. Direct link to Will Simon's post “What options are availabl...”
What options are available if H=0?
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(9 votes)
Answer
Duc Tran
Posted 3 years ago. Direct link to Duc Tran's post “How can we apply the seco...”
How can we apply the second partial derivative test for functions with more than 2 variables (like f(x,y,z))?
Button navigates to signup pageComment on Duc Tran's post “How can we apply the seco...”
(5 votes)
Answer
disha.bandy
Posted 6 years ago. Direct link to disha.bandy's post “How do I find the second ...”
How do I find the second partial derivatives for a function with 3 variables and how does this test work for that? Thanks :)
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(4 votes)
Answer
- John Smith
  Posted 2 years ago. Direct link to John Smith's post “You actually need to look...”
  You actually need to look at the eigenvalues of the Hessian Matrix, if they are all positive, then there is a local minimum, if they are all negative, there is a local max, and if they are of different signs, then there is a saddle.
  Button navigates to signup page
  (2 votes)
garcia.callie.88
Posted 6 years ago. Direct link to garcia.callie.88's post “What do you do if the sec...”
What do you do if the second partial derivative still has variables in it? How do you know if fxx or fyy is positive or negative?
Button navigates to signup pageButton navigates to signup page
(1 vote)
Answer
- Alexander Wu
  Posted 6 years ago. Direct link to Alexander Wu's post “If the second partial der...”
  If the second partial derivative is dependent on x and y, then it is different for different x and y. fxx(0, 0) is different from fxx(1, 0) which is different from fxx(0, 1) and fxx(1, 1) and so on. There's nothing wrong with that. You need to decide which point you care about and plug in the x and y values.
  
  Recall that was also the case with the second derivative test in single var calculus. You calculate the first or second derivative at some point.
  Comment on Alexander Wu's post “If the second partial der...”
  (4 votes)
Eva Xu
Posted 2 years ago. Direct link to Eva Xu's post “What if H>0 but fxx=0?”
What if H>0 but fxx=0?
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(1 vote)
Answer
- [name]
  Posted 2 years ago. Direct link to [name]'s post “This isn't possible Since...”
  This isn't possible
  Since fxy=fyx, (fxy)*(fyx) must be positive or 0. If fxx=0, then (fxx)*(fyy)=0. The determinant H is (fxx)*(fyy)- (fxy)*(fyx), so if fxx=0, then H cannot be greater than 0.
  Hope this helps :)
  Comment on [name]'s post “This isn't possible Since...”
  (4 votes)
marry shinota
Posted 2 years ago. Direct link to marry shinota's post “one claim is wrong, if p<...”
one claim is wrong, if p<-2 than p^2 will greater than 2*2
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(2 votes)
Answer
Colin Brankin
Posted 7 years ago. Direct link to Colin Brankin's post “Is the formula for the se...”
Is the formula for the second partial derivative test (fxx*fyy - (fxy)^2) just the determinant of the Hessian matrix learned earlier in this section?
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(2 votes)
Answer
- eirtanner
  Posted 2 years ago. Direct link to eirtanner's post “In a way, yes. It is a bo...”
  In a way, yes. It is a boiled-down version of the Hessian's determinant that excludes non-negative numerical factors.
  Comment on eirtanner's post “In a way, yes. It is a bo...”
  (1 vote)
Ali Qureshi
Posted 7 years ago. Direct link to Ali Qureshi's post “We were taught that the t...”
We were taught that the test actually was H = (f_12)^2 - (f_11)(f_22) and that if H > 0, the point is a saddle point and of course the opposite for local max/min. This contradicts what you have written here.
Button navigates to signup pageComment on Ali Qureshi's post “We were taught that the t...”
(0 votes)
Answer
- Martin Kajan
  Posted 7 years ago. Direct link to Martin Kajan's post “It's absolutelly OK. let...”
  It's absolutelly OK.
  lets say...
  fxx*fyy - (fxy)^2 > 0 meaning it's max/min.
  What you have provided here is this inequality*(-1). Proof:
  -(fxx*fyy) + (fxy)^2 < 0 is still a local maximum/minimum.
  Was my explanation clear enough?
  Button navigates to signup page
  (4 votes)
clara.vdw
Posted 7 years ago. Direct link to clara.vdw's post “Under '[Phrased using the...”
Under '[Phrased using the Hessian determinant]', we say that the second derivative test can be done using the determinant of the Hessian. This is only true for functions with input space in R2, right? With higher-dimensional input space, we need H to be positive definite. Correct?
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(1 vote)
Answer