If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

### Course: Multivariable calculus>Unit 3

Lesson 3: Optimizing multivariable functions

# Second partial derivative test example, part 1

A worked example of finding a classifying critical points of a two-variable function. Created by Grant Sanderson.

## Want to join the conversation?

• minor mistake you claim to have a grand total of three critical points while in fact you have 4
• I have the equation: x^2 + 6xy +10y^2 - 4y + 4. If I take the first derivative of x I end up with: 2x + 6y. What am I supposed to do with this? If I set it equal to zero I just keep ending up with x = -3y...?
• If you are looking for critical points, you will want to find the places where the tangent plane has zero slope. You will want to know where both partial df/dx and partial df/dy equal zero. In your example, you would calculate that partial df/dy is 6x +20y-4. Now you have two equations equal to zero with two variables. Just use algebra to solve the system of equations.
• What about inflection points? Where do they fit in in the gradient and the second partial derivative test?