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Show that the following functions $$f(x, y)=\frac{xy}{\sqrt{x^2+y^2}} \\ f(x, y)=\frac{x^2y}{x^4+y^2}$$ are differentiable at each point of the domain. Determine which of them is $C^1$.

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The domain of the functions is $\mathbb{R}^2 \setminus \{(0,0)\}$.

How can we show that the functions are differentiable at each point of the domain??

Do we have to find the partial derivatives??

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EDIT:

$$f(x, y)=\frac{xy}{\sqrt{x^2+y^2}}: \\ \frac{\partial{f}}{\partial{x}}=\frac{y\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2x}{x^2+y^2}=\frac{y(x^2+y^2)-x^2y}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{y^3}{\sqrt{x^2+y^2}(x^2+y^2)} \\ \frac{\partial{f}}{\partial{y}}=\frac{x\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2y}{x^2+y^2}=\frac{x(x^2+y^2)-xy^2}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{x^3}{\sqrt{x^2+y^2}(x^2+y^2)}$$

Since the partial derivatives are continuous at $\mathbb{R}^2 \setminus \{ (0,0)\}$, $f$ is differentiable.

Is this correct??

Could I improve something at the fomrmulation??

Mary Star
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    At each point of that domain each of the functions' derivatives of first order exist and are continuous = the function's differentiable at each point of that domain. – Timbuc Apr 02 '15 at 21:30
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    @MaryStar Check the definition? – Git Gud Apr 02 '15 at 22:07
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    @Mary: check whether your functions have continuous derivatives of first order. – Timbuc Apr 02 '15 at 22:10
  • Which is the difference to prove that a function is differentiable and to prove that a function is $C^1$ ??

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    In my book there is the following definition:

    Let $f: U \subset \mathbb{R}^n \rightarrow \mathbb{R}^m$. Let's suppose that all the partial derivatives $\frac{\partial{f_i}}{\partial{x_j}}$ of $f$ exist and are continuous at a region of a point $x \in U$. Then $f$ is differentiable in $x$.

    We say that a function of which the partial derivatives exist and are continuous are $C^1$. That means that each $C^1$ function is differentiable. @Timbuc

     – Mary Star Apr 02 '15 at 22:15
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    @MaryStar The paragraph you stated is part definition, part theorem. A $C^1$ function is a function whose first order partials are continuous. And there's a result that says that $C^1$ functions are differentiable. – Git Gud Apr 02 '15 at 22:17
  • And what is the definition that a function is differentiable ?? @GitGud – Mary Star Apr 02 '15 at 22:24
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    @MaryStar You should really check your notes. But you can find the definition here, at the end of the first part of the answer. – Git Gud Apr 02 '15 at 22:26
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    @MaryStar You understood it correctly. – Git Gud Apr 02 '15 at 22:28
  • With this definition it always stands that a differentiable function is also $C^1$, or not?? @GitGud Isn't there a difference between the two meanings?? – Mary Star Apr 02 '15 at 22:33
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    @MaryStar No, even with $f\colon \mathbb R\to \mathbb R$ there are well known counter examples. – Git Gud Apr 02 '15 at 22:35
  • So, at the exercise of my initial post do we show the differentiability using the limit defintion and to show that it is $C^1$ we show that the partial derivatives are continuous?? @GitGud – Mary Star Apr 02 '15 at 22:46
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    If they are $C^1$, they are differentiable, you don't need to use the definition for this. You do need to prove that the function is $C^1$, but that's just making the observation that it's the quotient of polynomials. – Git Gud Apr 02 '15 at 22:53
  • Do we say the following:$$f(x, y)=\frac{xy}{\sqrt{x^2+y^2}}: \ \frac{\partial{f}}{\partial{x}}=\frac{y\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2x}{x^2+y^2}=\frac{y(x^2+y^2)-x^2y}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{y^3}{\sqrt{x^2+y^2}(x^2+y^2)} \ \frac{\partial{f}}{\partial{y}}=\frac{x\sqrt{x^2+y^2}-xy\frac{1}{2\sqrt{x^2+y^2}}2y}{x^2+y^2}=\frac{x(x^2+y^2)-xy^2}{\sqrt{x^2+y^2}(x^2+y^2)}=\frac{x^3}{\sqrt{x^2+y^2}(x^2+y^2)}$$ Since the partial derivatives are continuous at $\mathbb{R}^2 \setminus { (0,0)}$, $f$ is $C^1$. That implies that $f$ is differentiable. $$$$ Is this correct?? @GitGud – Mary Star Apr 02 '15 at 23:01
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    @MaryStar It is. – Git Gud Apr 02 '15 at 23:03
  • Could I improve something at the formulation?? @GitGud – Mary Star Apr 02 '15 at 23:05
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    @MaryStar It's fine in my opinion, I would expect full marks. – Git Gud Apr 02 '15 at 23:07
  • If we had a function that wasn't $C^1$ how would we show that it is differentiable?? Using the limit definition?? Or is there aso an other way?? – Mary Star Apr 03 '15 at 00:07

1 Answers1

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Since $(0,0)$ isn't in the domain, then we don't need to worry about that point. The necessary and sufficient conditions for multi-variable differentiation are the existence of all the partials, and their continuity at a given point.

Emir Šemšić
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