Is this f(x,y) differential at (1,2)?

In summary: Actually my previous suggestion wasn't good enough. Try and write your numerator in the form (x-a)*(x-1)^2+(y-b)*(y-2)^2+(x-c)*(x-1)*(y-2). Expand that and find the constants a, b and c that work. Now it's kind of like your polar coordinate example if you think of |x-1|~|y-2|~r. The denominator is degree r, the numerator is degree r^2.
  • #1
Jamin2112
986
12

Homework Statement



Let f(x,y)=x3 + x2y + y3. Show that f is differentiable at (1,2).

Homework Equations



A function f(x,y) is differentiable at a point (x0,y0) if the partial derivatives of f(x,y) are defined at (x0,y0) and if the following limiting condition holds:

lim(x,y)-->(x0,y0) [ f(x,y) - f(x0,y0) - ∂f/∂x|(x0,y0)(x-x0) - ∂f/∂y|(x0,y0)(y-y0) ] / √( (x-x0)2 + (y-y0)2) = 0.

The Attempt at a Solution



∂f/∂x = 3x2 + 2xy ===> ∂f/∂x|(1,2) defined

∂f/∂y = 2x2 + 3y2 ===> ∂f/∂y|(1,2) defined

But I'm having trouble with making lim(x,y)-->(1,2) = 0.

I got it all the way to

lim(x,y)-->(1,2) = (x3+x2y+y3+22 -7x-13y)/√((x-1)2+(x-2)2).

I need to get rid of the denominator. I thought about rationalizing the denominator, but I don't that'll get me anywhere.

suicide-bear-didnt-even-leave-a-not.jpg


Suggestions welcome.
 
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  • #2
In case my typing wasn't clear, here's a link to the problem: http://www.math.washington.edu/~sullivan/3262_sp10.pdf"
 
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  • #3
You know in your gut that it's true that it's differentiable, right? So you know that you just somehow MUST be able to factor the numerator as (x-1)*g(x,y)+(y-2)*h(x,y) for some polynomials g(x,y) and h(x,y), right? Divide the numerator by x-1. Take the remainder from that and divide by y-2. Nice decapitation photo. I sympathize.
 
  • #4
Dick said:
You know in your gut that it's true that it's differentiable, right? So you know that you just somehow MUST be able to factor the numerator as (x-1)*g(x,y)+(y-2)*h(x,y) for some polynomials g(x,y) and h(x,y), right? Divide the numerator by x-1. Take the remainder from that and divide by y-2. Nice decapitation photo. I sympathize.
Hmmm...

So I have (x-1)P1(x,y) + (y-2)P2(x,y) = x3 + x2y + y3 + 22 - 7x - 13y.

But I don't understand your method of solving this. (I'm not looking for a straight-up answer, just a little more detailed explanation on how to simplify this)

Also, maybe some help with part (b).

Part (b) reads:

Check if f(x,y) = (xy2-y3)/(x2+y2) is differentiable at (0,0) using polar coordinates.


My attempt:

Obviously, since the numerator is homogeneous of degree 3, and the denominator of degree 2, we will have limr-->0 r * h(ø) = 0, because h(ø) is just kind of there but always -1 ≤ h(ø) ≤ 1 while r sends it to zero.

Here's the thing, though. I also need to show that "the partial derivatives of f(x,y) are defined at (x0,y0)" (that from part (a)). In this case, of course, I'm using f(rcosø,rsinø) instead of f(x,y). But in dealing with the limit in polar coordinates, I'm only supposed to worry about r. How can their be partial derivatives?
 
  • #5
Every polynomial in x and y is differentiable for all (x, y).
 
  • #6
Jamin2112 said:
Hmmm...

So I have (x-1)P1(x,y) + (y-2)P2(x,y) = x3 + x2y + y3 + 22 - 7x - 13y.

But I don't understand your method of solving this. (I'm not looking for a straight-up answer, just a little more detailed explanation on how to simplify this)

Actually my previous suggestion wasn't good enough. Try and write your numerator in the form (x-a)*(x-1)^2+(y-b)*(y-2)^2+(x-c)*(x-1)*(y-2). Expand that and find the constants a, b and c that work. Now it's kind of like your polar coordinate example if you think of |x-1|~|y-2|~r. The denominator is degree r, the numerator is degree r^2.
 

What does it mean for a function to be differential?

For a function to be differential at a point, it means that the function is differentiable at that point. This means that the function has a well-defined derivative at that point, which represents the instantaneous rate of change of the function at that point.

How do you determine if a function is differential at a given point?

To determine if a function is differential at a given point, we can use the definition of differentiability. This involves checking if the limit of the difference quotient of the function exists as the independent variables approach the given point. If the limit exists, the function is differentiable at that point. Alternatively, we can also check if the partial derivatives of the function exist at that point.

Is there a difference between a function being differentiable and differential?

No, there is no difference between a function being differentiable and differential. These terms are often used interchangeably to describe a function that has a well-defined derivative at a given point.

Can a function be differential at some points but not others?

Yes, it is possible for a function to be differential at some points but not others. This depends on the behavior of the function and the specific points being evaluated. For example, a function may be differentiable at most points but not at the points where it has a sharp corner or discontinuity.

What are the applications of determining if a function is differential at a given point?

Determining if a function is differential at a given point is important in many areas of mathematics and science. It allows us to understand the behavior of the function and make predictions about its rate of change. This is useful in fields such as physics, engineering, economics, and more.

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