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I'm given a function like $(x^2+y^2)sin(1/\sqrt{x^2+y^2})$ that's $0$ at $(x,y)=0$, and the function is obviously not of class $C^1$, is there any way to prove that it's differentiable other than the limit method showing $\exists B,lim_{h\rightarrow 0}(f(a+h)-f(a)-Bh)/||h||=0$ ?

Otherwise, is there any particular (general) tricks to proving functions like this are differentiable through the limit method?

Note: Proving a scalar function is differentiable at the origin but that its partial derivatives are not continuous at that point. doesn't give everything I'm looking for, because I want to show differentiability everywhere.

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I'm not super sure if I understand your question completely, because the continuity of partial derivatives can be extended to the entire space (and not just a single point). Perhaps it will be helpful to look at the actual statement of the theorem, which we discuss below.

One can use the following theorem from Munkres' Analysis on Manifolds that is sufficient for differentiation but not necessary:

Theorem 6.2. Let $A$ be open in $\mathbb{R}^m$. Suppose that the partial derivatives $D_j f_i (x)$ of the component functions of $f$ exist at each point $x$ of $A$ and are continuous on $A$. Then $f$ is differentiable at each point of $A$.

In your case, take $\frac{d}{dx}, \frac{d}{dy}$ and see if they are continuous. If so, then your function $f$ is differentiable at each point.

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  • $\begingroup$ I'm aware of the theorem, but in this case, aren't the partial derivatives discontinuous at 0, having periods that approach zero? $\endgroup$ – George Nov 10 '16 at 7:15
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For this particular function yes, there is one helpful trick:

Notice that the function $f(x,y)=\frac{x^2+y^2}{\sin(1/\sqrt{x^2+y^2})}$ is actually a function of $x^2+y^2$; i.e. $f(x,y)=g(x^2+y^2)$, where $$ g(r):=\frac{r}{\sin(1/\sqrt{r})}. $$ How does this help? The crux of the matter is that $$ (x,y)\to(0,0)\hspace{.7cm}\Longleftrightarrow\hspace{.7cm}x^2+y^2\to0, $$ which, along with the previous observation, leads to $$ \lim_{(x,y)\to(0,0)} f(x,y)=\lim_{x^2+y^2\to0}f(x,y)=\lim_{x^2+y^2\to0}g(x^2+y^2)=\lim_{r\searrow 0}g(r). $$ This allows us to deal with a hopefully easier limit in the more familiar one-variable case. In this particular example $g$ is highly oscillating when $r\searrow0$ so the limit won't exist, and then $f$ won't be differentiable at $(0,0)$.

Proving that a multivariate function has a limit can be a tedious work. When the "partial derivatives are $C^1$"--test does not apply one needs to get creative, and the particular tricks at hand might or might not work. In my experience this is more of an ad-hoc situation. However you probably will find helpful the answers given in this MSE post.

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You say the function is obviously not $C^1.$ Why is that obvious? I can see that on the $x$-axis, we have the function $x^2\sin (1/|x|).$ And I know from one variable experience that the partial derivative with respect to $x$ will not be continuous at $(0,0).$ Is that what you meant?

Because actually your function belongs to $C^\infty(\mathbb R^2\setminus \{(0,0)\}).$ (In fact it's real analytic in that domain.) Why? Because compositions of $C^\infty$ functions are $C^\infty.$ And on $R^2\setminus \{(0,0)\},$ we are looking at the composition of $t^2\sin (1/|t|),$ which is in $C^\infty(R\setminus \{0\}),$ with the function $(x^2+y^2)^{1/2}, $ which is in $C^\infty(\mathbb R^2\setminus \{(0,0)\}).$

Thus your function is differentiable everywhere in $\mathbb R^2\setminus \{(0,0)\}.$ So no, you do not need to go throuh the limit procedure everywhere. Far from it: You only have to check differentiability at $(0,0).$ You appear to know how to do that, so I'll leave off here.

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