Partial derivatives of function log(x^2+y^2)

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SUMMARY

The discussion centers on the partial derivatives of the function f(x,y) = log(x² + y²). The correct second partial derivatives are given as ∂²f/∂x² = (y² - x²) / (x² + y²)² and ∂²f/∂y² = (x² - y²) / (x² + y²)². The conclusion drawn is that ∂²f/∂x² = -∂²f/∂y², which is valid due to the nature of partial differentiation where the variable not being differentiated is treated as a constant. A common misconception is clarified: the roles of x and y are not identical in the context of differentiation.

PREREQUISITES
  • Understanding of partial derivatives in multivariable calculus
  • Familiarity with logarithmic functions and their properties
  • Knowledge of the chain rule in differentiation
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the derivation of partial derivatives for multivariable functions
  • Learn about the implications of treating variables as constants during differentiation
  • Explore the applications of second partial derivatives in optimization problems
  • Review the properties of logarithmic functions in calculus
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Students studying calculus, particularly those focusing on multivariable functions, educators teaching advanced mathematics, and anyone seeking to deepen their understanding of partial differentiation.

Chromosom
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Homework Statement


I have got a question concerning the following function:

f(x,y)=\log\left(x^2+y^2\right)​

Partial derivatives are:

\frac{\partial^2f}{\partial x^2}=\frac{y^2-x^2}{\left(x^2+y^2\right)^2}​

and

\frac{\partial^2f}{\partial y^2}=\frac{x^2-y^2}{\left(x^2+y^2\right)^2}​

The conclusion is that the following equation is right:

\frac{\partial^2f}{\partial x^2}=-\frac{\partial^2f}{\partial y^2}​

But I can not understand, how can it be possible. The role of x and y variables are exactly the same, then why derivatives are not the same?

Sorry for my English - it is my second language. I am from Poland.
 
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Chromosom said:

Homework Statement


I have got a question concerning the following function:

f(x,y)=\log\left(x^2+y^2\right)​

Partial derivatives are:

\frac{\partial^2f}{\partial x^2}=\frac{y^2-x^2}{\left(x^2+y^2\right)^2}​

and

\frac{\partial^2f}{\partial y^2}=\frac{x^2-y^2}{\left(x^2+y^2\right)^2}​

The conclusion is that the following equation is right:

\frac{\partial^2f}{\partial x^2}=-\frac{\partial^2f}{\partial y^2}​

But I can not understand, how can it be possible. The role of x and y variables are exactly the same, then why derivatives are not the same?

Sorry for my English - it is my second language. I am from Poland.

There's a factor of 2 missing in all your second derivatives.

The result is exactly as you'd expect. The variable you're differentiating with respect to, matters. If it's x, then y is treated as a constant, and vice versa. So if the "active" variable is leading in the numerator in one derivative, the same should apply in the other. It's just that the "active" variable is x in one case and y in the other, and the other variable acts like a constant.
 
Chromosom said:

Homework Statement


I have got a question concerning the following function:

f(x,y)=\log\left(x^2+y^2\right)​

Partial derivatives are:

\frac{\partial^2f}{\partial x^2}=\frac{y^2-x^2}{\left(x^2+y^2\right)^2}​

and

\frac{\partial^2f}{\partial y^2}=\frac{x^2-y^2}{\left(x^2+y^2\right)^2}​

The conclusion is that the following equation is right:

\frac{\partial^2f}{\partial x^2}=-\frac{\partial^2f}{\partial y^2}​

But I can not understand, how can it be possible. The role of x and y variables are exactly the same, then why derivatives are not the same?

Sorry for my English - it is my second language. I am from Poland.

How did you get those partial derivatives? They are wrong.


P.S. There's nothing wrong with your English, and even if there were, there is nothing to apologise for.
 

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