# Stationary points (roots) of partially derivated function

## 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

## 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?

## Answers and Replies

Dick
Science Advisor
Homework Helper
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 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?

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?

Dick
Science Advisor
Homework Helper
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?

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?

Dick
Science Advisor
Homework Helper
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.