Question

Theorem: Let \(A\subset \mathbb R^n\) be open and \(f:A\subset \mathbb R^n\rightarrow \mathbb R^m\). Suppose \(f=f_1,....,f_n\). If each of the partials \(\partial f_j/\partial x_i\) exists and is continous on A, then f is differentiable on A

Answer 1

what is the question

Answer 2

Use that theorem to prove that the f(x,y) is differentiable at (0,0)
\[f(x,y) =\dfrac{(xy)^2}{\sqrt{x^2+y^2}} \] where \((x,y)\neq (0,0)\)

Answer 3

and f(x,y) =0 if (x,y) =(0,0)

Answer 4

So to find partial derivatives, just take the function w.r.t x and then w.r.t y.

Answer 5

Find what \(z_{x}\) and \(z_y\) are when \(f(x,y) = z=\dfrac{(xy)^2}{(x^2+y^2)^{1/2}}\)

Answer 6

and \(z_x \iff \dfrac{dz}{dx}\) and \(z_y \iff \dfrac{dz}{dy}\)

Answer 7

I know how to find partial derivative, what is next? need prove each of them is continuous on R^n? how?

Answer 8

Ok, when you have found the partial derivative, plug in your coordinate point,. in this case, (0,0) to find your partial.

Answer 9

Got what you mean; give me time to solve it, then check or guide more, please.

Answer 10

I got \(\dfrac{\partial f}{\partial x}=\dfrac{y^2x^3+2xy^4}{(x^2+y^2)^{3/2}}\)

\(\dfrac{\partial f}{\partial y}=\dfrac{x^2y^3+2yx^4}{(y^2+x^2)^{3/2}}\)

Answer 11

since f(x,y) is a symmetric function, we can work on partial w.r.t x and then change the variable to y to get both.
From the proof above, \(\dfrac{\partial f}{\partial x}\) exist and continuous everywhere but (0,0)

Answer 12

Oh, I misunderstand the theorem. It says partial exist and f itself is continuous, not partial continuous.
hence we need prove \(lim_{(x,y)\rightarrow (0,0)}\dfrac{(xy)^2}{\sqrt{x^2+y^2}}\) continuous

Answer 13

It's easier if we use polar

Answer 14

No, you were right. I'm reading about it right now:
http://mathinsight.org/differentiability_multivariable_theorem

Answer 15

ah!!:(

Answer 16

lol

Answer 17

Let define partial w.r.t x = h(x,y) , then need show \(lim_{(x,y)\rightarrow (0,0)} h(x,y) = h(0,0)\) hence h(x,y) is continuous, right?

Answer 18

Seems right to me. I need to do some more reading to make sure I've covered the subject before I can really help.

Answer 19

rely on you :)

Answer 20

Alright, sorry it took so long, I wasn't focused yesterday..
Anyway, I did some reading and gathered information from different resources.
However, I'm new to the subject so I suggest verifying what I'm saying.
Here is how I see it:

Our function is:
$$
f(x, y) = \begin{cases}
\frac{(xy)^2}{\sqrt{x^2 + y^2}}, & x \neq 0 \\
0, & x = 0
\end{cases}
$$
First we have to define our partial derivative with respect to x. Let it be \(D_x(x,y)\).
For \((x,y) \neq (0,0)\) we can simply say:
$$
D_x(x,y) = \frac{x^3 y^2 + 2xy^4}{(x^2 + y^2)^\frac{3}{2}}
$$But we still have to define our function for \((x,y) = (0,0)\).
We can do it the old-fashioned way and use the derivative limit.
$$
y = 0 \\
f(x, 0) = \frac{(xy)^2}{\sqrt{x^2 + y^2}} = \frac{(x \cdot 0)^2}{\sqrt{x^2 + 0^2}} = \frac{0}{\sqrt{x^2}} = 0\\
D_x(0,0) = \lim_{\Delta x \to 0}\frac{f(x + \Delta x, y) - f(x,y)}{\Delta x} = \\
= \lim_{\Delta x \to 0}\frac{f(0 + \Delta x, 0) - f(0,0)}{\Delta x} = \lim_{\Delta x \to 0}\frac{0 - 0}{\Delta x} = 0
$$Because no matter what value we get for x, the function's value when y=0 is always 0.
In order for a function to be continuous at a given point, there should be a limit at that point that approaches the function's value of the point.
That is why we had to define the partial derivative at (0,0) here.
Now we want to show that \(D_x(x,y)\) has a limit at (0,0) and that limit approaches \(D_x(0,0)\) in order to show it is continuous:
\[ \lim_{(x,y) \to (0,0)} \frac{x^3 y^2 + 2xy^4}{(x^2 + y^2)^\frac{3}{2}} \stackrel{?}{=} D_x(0,0)
\]We can do so by converting to polar coordinates.
Say that for every (x,y) we set (\(\theta, r\)) such that:
$$
x = r \cdot cos(\theta) \\
y = r \cdot sin(\theta) \\
$$Now \(\theta\) corresponds to the direction we're taking the limit and r will be the distance from the origin of the examined point.
If we can show that for any possible direction \(\theta\) we get the same limit when approaching r to 0 and that limit equals to \(D_x(0,0)\) , then the function is continuous at the (0,0).
$$
\lim_{(x,y) \to (0,0)}D_x(x,y) = \lim_{(x,y) \to (0,0)} \frac{x^3 y^2 + 2xy^4}{(x^2 + y^2)^\frac{3}{2}} = \\
= \lim_{r \to 0} \frac{(r \cdot cos(\theta))^3 (r \cdot sin(\theta))^2 + 2(r \cdot cos(\theta))(r \cdot sin(\theta))^4}{((r \cdot cos(\theta))^2 + (r \cdot sin(\theta))^2)^\frac{3}{2}} = \\
= \lim_{r \to 0} \frac{ r^5 (cos^3(\theta) \cdot sin^2(\theta)) + 2r^5(cos(\theta) \cdot sin^4(\theta))}{(r^2[cos^2(\theta) + sin^2(\theta)])^\frac{3}{2}} = \\
= \lim_{r \to 0} \frac{ r^5 (cos^3(\theta) \cdot sin^2(\theta) + 2cos(\theta) \cdot sin^4(\theta))}{r^3} = \\
= \lim_{r \to 0} r^2 (cos^3(\theta) \cdot sin^2(\theta) + 2cos(\theta) \cdot sin^4(\theta)) = 0 = D_x(0,0)
$$Because the trigonometric functions are bounded functions and therefore when r approaches 0 this whole thing approaches 0.
Since the function's limit approaches the function's value at (0,0) it is continuous there.

We can do the same for y and it will look exactly (really) the same.
Hope it helps \( \Large ☺\)

Answer 21

It's a big help :)

Answer 22

I like the polar part. Thank you so much.