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

[Jamin2112]

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.

Suggestions welcome.

 

[Jamin2112]

In case my typing wasn't clear, here's a link to the problem: http://www.math.washington.edu/~sullivan/3262_sp10.pdf"

 

[Dick]

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.

 

[Jamin2112]

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 lim_r-->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?

 

[HallsofIvy]

Every polynomial in x and y is differentiable for all (x, y).

 

[Dick]

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.