Understanding Partial Derivatives of x^2 + y^2 < 1

• emergentecon
In summary, the conversation discusses finding the partial derivatives of a function where x^2 + y^2 < 1. The solution is given as @f/@x = 2x = 0 and @f/@y = 2y + 1 = 0. The poster initially had a question about the solution, but later realized their error and no longer needed to delete the question.

Homework Statement

x^2 + y^2 < 1

Find the partial derivatives of the function.

x^2 + y^2 < 1

The Attempt at a Solution

@f/@x = 2x = 0
@f/@y = 2y = 0

4. Their solution

@f/@x = 2x = 0
@f/@y = 2y + 1 = 0

5. My Problem

I don't see how / why they get 2y + 1 for the partial of y?

Issue is resolved, there was an error in reading the text.
I cannot see how to delete the question though?

Questions about differentiation should be posted in the Calculus & Beyond section.

Just because a question has been answered it is not ordinarily deleted.

1. Given the elementary nature of the question, why is Calculus & Beyond Homework not the correct section?
2. It is not a function of deleting an answered question. I misunderstood something I was reading, and thereby posted an erroneous question.

emergentecon said:
1. Given the elementary nature of the question, why is Calculus & Beyond Homework not the correct section?
It IS the correct section. You initially posted it in the Precalculus section. I moved it.
emergentecon said:
2. It is not a function of deleting an answered question. I misunderstood something I was reading, and thereby posted an erroneous question.
You can post the corrected question here in this thread.

Ok . . . mea culpa for the section error.

Mark44 said:
It IS the correct section. You initially posted it in the Precalculus section. I moved it.

You can post the corrected question here in this thread.
Once I realized my error, I no longer had a problem with the question.
Hence wanting to delete it. But I see how it works, thank you.