Stationary points (roots) of partially derivated function

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Homework Help Overview

The discussion revolves around finding stationary points of the function f(x,y) = ln(x+y) - x² - y², focusing on the calculation of partial derivatives and the conditions under which these derivatives equal zero. The problem is situated within the context of multivariable calculus.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to compute the partial derivatives of the function but expresses uncertainty about the logarithmic component. They question the definition of roots in this context and how to set the derivatives to zero.
  • Some participants confirm the correctness of the derivatives and clarify that stationary points occur where both derivatives equal zero simultaneously.
  • Another participant suggests a potential relationship between x and y, proposing that if both derivatives equal zero, then x might equal y.
  • Further discussion arises regarding the implications of solving for x and y, particularly in relation to quadratic equations and their roots.

Discussion Status

The conversation is active, with participants providing guidance on the interpretation of derivatives and the conditions for stationary points. There is an exploration of multiple interpretations regarding the relationship between x and y, and the discussion remains open without a definitive conclusion.

Contextual Notes

Participants are navigating through the complexities of derivatives and their implications in a multivariable context. There is also a mention of a separate limit problem, indicating a broader scope of inquiry within the thread.

brollysan
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Homework Statement



f(x,y)= ln(x+y) -x^2 - y^2

1. Find the partially derivatives
2. Find the stationary points (roots) of the function

Homework Equations





The Attempt at a Solution



Quite simple, except i don't know what to do with the ln part, this is my attempt tho

f'x= 1/(x+y) -2x
f'y= 1/(x+y) -2y

Is this right? I am not sure what the derivative of ln (x+ c) is c= constant

2. Roots= where the function crosses the x-line right? So I simply set f'x=0 and solve it to get the root?
 
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f'x and f'y are correct. The derivative of ln(f(x)) is f'(x)/f(x). They then want values of x and y where BOTH f'x and f'y equal zero simultaneously. These are the stationary points. Can you find them?
 
If they both equal zero then they must both equal each other right? Then that means x=y?

F'x=0 = 1/(2x) -2x => 4x^2 =1 => x= 1/4

And since x = y then they must both have a root of 1/4...am i right?
 
While i am at it i also need help with a limit problem.

The problem, all variables etc

f(x)= x^2 + 2x

find

lim x->a [f(x) - f(a)]/ (x-a)

The attempt at a solution

The correct answer is 2a +2

Factoring etc i got this far:

[(x-a)(x+a) -2a -2x]/(x-a)

Now i can write this as [(x-a)(x+a)]/(x-a) + -2a-2x / (x-a)

Leaving me with x+a -2a-2x/(x-a) I am stuck with this, how do make it so the last part simply becomes 2?
 
brollysan said:
If they both equal zero then they must both equal each other right? Then that means x=y?

F'x=0 = 1/(2x) -2x => 4x^2 =1 => x= 1/4

And since x = y then they must both have a root of 1/4...am i right?

No. x^2=1/4, not x=1/4. And remember quadratic equations generally have two roots. What are the (x,y) pairs that satisfy this?
 
brollysan said:
While i am at it i also need help with a limit problem.

The problem, all variables etc

f(x)= x^2 + 2x

find

lim x->a [f(x) - f(a)]/ (x-a)

The attempt at a solution

The correct answer is 2a +2

Factoring etc i got this far:

[(x-a)(x+a) -2a -2x]/(x-a)

Now i can write this as [(x-a)(x+a)]/(x-a) + -2a-2x / (x-a)

Leaving me with x+a -2a-2x/(x-a) I am stuck with this, how do make it so the last part simply becomes 2?

Surely you can simplify (-2a-2x)/(x-a). (x-a) is a factor of the numerator.
 

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