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

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SUMMARY

The discussion focuses on finding the partial derivatives of the function defined by the inequality x^2 + y^2 < 1. The correct partial derivatives are established as ∂f/∂x = 2x and ∂f/∂y = 2y + 1. A misunderstanding regarding the notation led to confusion, which was clarified by community members. The original poster acknowledged their error in categorizing the question and expressed gratitude for the resolution.

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emergentecon
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Homework Statement



x^2 + y^2 < 1

Find the partial derivatives of the function.

Homework Equations



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?
 
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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.
 

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