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Let $p \geq 1$ and $f: \mathbb{R^2} \to \mathbb{R}$ defined as

$$f(x) = \begin{cases} (\sin \|x\|)^p \cos \frac{1}{\|x\|}, & \quad \text{if } \|x\| \not= 0 \\ 0, & \quad \text{if } \|x\| = 0 \\ \end{cases} $$

Show that $f$ is differentiable at point $x \not= (0,0)$

I don't know if there exist a shorter way to solve this problem, but here what I want to explore.

I want to find the partial derivatives of $h(x) = (\sin \|x\|)^p \cos \frac{1}{\|x\|}$ to get the gradient at the point $\bar{x}=(x_1,y_1) \not= (0,0)$. From this time, I can use the gradient at $\bar{x}$ to show that $f$ is differentiable at point $x \not= (0,0)$ in using the definition of differentiability (with the linear function $\nabla f(\bar{x}) \cdot (x-\bar{x})$).

$$\frac{\partial f}{\partial x}(x_1,y_1) = \lim_{h \to 0} \frac{f(x_1+h,y_1)-f(x_1,y_1)}{h} = \lim_{h \to 0} \frac{(\sin \|(x_1+h,y_1)\|)^p\cos \frac{1}{\|(x_1+h,y_1)\|}-\sin\|(x_1,y_1)\|)^p\cos \frac{1}{\|(x_1,y_1)\|}}{h}$$

I know the explicite formula for each partial derivative, but I have to show with the definition of a partial derivative. However, I am blocked to show $\frac{\partial f}{\partial x}(x_1,y_1)$ with the limit.

Is anyone could help me to find the explicite partial derivative of $\frac{\partial f}{\partial x}(x_1,y_1)$? Is there a easier way to show that $f$ is differentiable at point $x \not= (0,0)$?

P.S. Please, don't try to use a very specific analysis theory; I am only an undergraduate student (bachelor).

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If you want to analyze differentiability away from $(0,0)$, you need not worry about the $f(0,0)$ part, just about the formula. In general, writing $x=(x_1,x_2)$ (I'm using notation different from your $(x_1,y_1)$ to make indexing easier, so I can compute all the partial derivatives at once), we have that: $$\frac{\partial}{\partial x_i}(\|x\|) = \frac{x_i}{\|x\|}, \quad \forall\,x \neq 0.$$So: $$\begin{align} \frac{\partial}{\partial x_i}\left((\sin \|x\|)^p\cos\left(\frac{1}{\|x\|}\right)\right)&= \frac{\partial}{\partial x_i}((\sin\|x\|)^p)\cos\left(\frac{1}{\|x\|}\right)+(\sin\|x\|)^p\frac{\partial}{\partial x_i}\left(\cos\left(\frac{1}{\|x\|}\right)\right) \\ &= p(\sin \|x\|)^{p-1}\cos\left(\frac{1}{\|x\|}\right)\frac{x_i}{\|x\|}+(\sin\|x\|)^p \sin\left(\frac{1}{\|x\|}\right)\frac{1}{\|x\|^2}\frac{x_i}{\|x\|} \\ &= p(\sin \|x\|)^{p-1}\cos\left(\frac{1}{\|x\|}\right)\frac{x_i}{\|x\|}+(\sin\|x\|)^p \sin\left(\frac{1}{\|x\|}\right)\frac{x_i}{\|x\|^3} ,\end{align}$$which are continuous functions of $x$ away from $(0,0)$. And if a function has continuous partial derivatives at a point, then it is differentiable at that point.


Here I assume that $\|(x_1,x_2)\| = \sqrt{x_1^2+x_2^2}$ is the euclidean norm.

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  • $\begingroup$ In your opinion, is a professor of analysis may be required to use the definition of the partial derivative to calculate $\frac{\partial f}{\partial x}(x_1,y_1)$? If so, is that long and painful exercise in the case that concerns me? $\endgroup$ – Taj Mohamed Bandalandabad Feb 13 '16 at 3:09
  • $\begingroup$ I guess it depends on how far the lecturer already went with the theory there. I agree that computing the partial derivatives of that function by definition is very painful. If you can use the chain and product rules, by all means, don't restrain yourself. $\endgroup$ – Ivo Terek Feb 13 '16 at 3:11
  • $\begingroup$ Thank for your solution BTW! In your opinion, is there a easier way to show that $f$ is differentiable at point $x \not= (0,0)$ or this way use is the best? $\endgroup$ – Taj Mohamed Bandalandabad Feb 13 '16 at 3:14
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    $\begingroup$ The usual ways to check differentiability are to see that the partial derivatives are continuous at the point (I did that above), or computing the partial derivatives ate the point and plugging them at that ugly limit (see "differentiability in higher dimensions"). I can't see an easier way right now.. $\endgroup$ – Ivo Terek Feb 13 '16 at 3:35

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