Stationary points (roots) of partially derivated function

In summary: So you can simplify it to what?In summary, in this conversation, the speaker asks for help finding the partial derivatives and stationary points of a function containing a logarithmic term. They provide their attempt at a solution and ask for confirmation. The expert summarizer explains that the derivatives are correct and guides the speaker to find the values of x and y where both derivatives equal zero. The expert also provides assistance with a limit problem, guiding the speaker to simplify the expression to find the correct answer.
  • #1
brollysan
27
0

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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  • #2
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?
 
  • #3
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?
 
  • #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?
 
  • #5
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?
 
  • #6
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.
 

1. What are stationary points of a partially derivated function?

Stationary points are points on a graph where the derivative of a function is equal to zero. In other words, they are points where the slope of the tangent line is horizontal and the function's rate of change is temporarily zero.

2. How do you find stationary points of a partially derivated function?

To find stationary points, you need to take the partial derivative of the function with respect to each independent variable and set them equal to zero. Then, solve the resulting system of equations to find the values of the independent variables at the stationary points.

3. What do stationary points tell us about a partially derivated function?

Stationary points provide important information about the behavior of a function. They can indicate the maximum or minimum values of the function, as well as points of inflection where the function changes from increasing to decreasing or vice versa.

4. How do you determine if a stationary point is a maximum, minimum, or point of inflection?

To determine the type of stationary point, you can use the second derivative test. Plug the values of the independent variables at the stationary point into the second derivative of the function. If the result is positive, the stationary point is a minimum. If it is negative, the point is a maximum. If the second derivative is zero, the stationary point is a point of inflection.

5. Can a partially derivated function have multiple stationary points?

Yes, a function can have multiple stationary points. These points can be local (only applicable to a specific portion of the graph) or global (applicable to the entire graph). It is important to consider all stationary points when analyzing the behavior of a function.

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