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Stationary points (roots) of partially derivated function

  1. Apr 2, 2008 #1
    1. The problem statement, all variables and given/known data

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

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

    2. Relevant equations



    3. The attempt at a solution

    Quite simple, except i dont 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?
     
  2. jcsd
  3. Apr 2, 2008 #2

    Dick

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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?
     
  4. Apr 2, 2008 #3
    If they both equal zero then they must both equal eachother 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?
     
  5. Apr 2, 2008 #4
    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?
     
  6. Apr 2, 2008 #5

    Dick

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    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?
     
  7. Apr 2, 2008 #6

    Dick

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