Find partial derivatives of $f:\mathbb{R}^n\rightarrow\mathbb{R}$ $$f(x)=\|x\|^\alpha$$ outside of $(0,0)$ when $\alpha\in\mathbb{R}$. What values does $\alpha$ have to take for the partial derivatives to also exist at $(0,0)$?
I only know that for the partial derivatives to exist at a point $a$, there has to be a limit $$\lim_{t\to0}\frac{f(a+te_k)-f(a)}{t}$$
but I don't know how to use it. I'm also not sure how to find partial derivatives without including all the points.
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$$f(x)=\left(\sum_{i=1}^n x_i^2\right)^{\alpha/2},$$ consider $x_k$ as the variable and the other components as constants.
Then using the usual differentiation rules,
$$\frac{\partial f(x)}{\partial{x_k}}=\frac\alpha2\,2x_k\,\left(\sum_{i=1}^n x_i^2\right)^{\alpha/2-1}=\alpha x_k(f(x))^{1-2/\alpha}.$$
(In the summation, only the term $x_k^2$ contributes.)
In particular,
for $\alpha=-1$, $-x_kf^3(x)$,
for $\alpha=1$, $\dfrac{x_k}{f(x)}$, and
for $\alpha=2$, $2x_k$.
Hint: $\Vert x \Vert = \sqrt{x_1^2 + x_2^2 + \cdots + x_n^2} = (x_1^2 + x_2^2 + \cdots + x_n^2)^{\frac{1}{2}}$, so $\Vert x \Vert^\alpha = (x_1^2 + x_2^2 + \cdots + x_n^2)^\frac{\alpha}{2}$. This is a nice differentiable function in all the $x_i$, except possibly at $0$.