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For

$$ f(x) = |x|^\alpha \quad x \in \mathbb{R}^n \setminus \{0\} $$

the first partial derivative is: $$ \begin{align*} \frac{\partial f}{\partial x_i} (x) = \frac{\partial\left( \sum_{k=1}^n x_k^2 \right)^{\frac{\alpha}{2}}}{\partial x_i} &= \frac{\partial}{\partial x_i}(x_1^2 + x_2^k + \cdots + x_i^2 + \cdots + x_n^2)^{\frac{\alpha}{2}} \\ &= \alpha x_i \left(\sum_{k=1}^n x_k^2 \right)^{\frac{\alpha}{2} -1} = \alpha x_i (f(x))^{1 - \frac{2}{\alpha}} \end{align*} $$

How can I get the second order partials i.e.

$$ \frac{\partial f}{\partial x_i \partial x_j} \quad \forall i,j \in \{1, \ldots, n\} $$

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    $\begingroup$ You must correct your notation about the indices. $\endgroup$ – John M-D94 Jun 7 '20 at 22:15
  • $\begingroup$ You're right. Thanks :) $\endgroup$ – millionmilesaway Jun 7 '20 at 22:30
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$\frac{\partial f(x)}{\partial x_{i}\partial x_{j}} = \alpha(\sum_{k=1}^{k=n} x^{2}_{k})^{\frac{\alpha}{2}-1}\frac{\partial x_{i}}{\partial x_{j}} + 2\alpha x_{i}x_{j}(\frac{\alpha}{2}-1)(\sum_{k=1}^{k=n} x^{2}_{k})^{\frac{\alpha}{2}-2}$

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You have correctly computed $$\frac{\partial f}{\partial x_i}=\alpha x_i\left(f\left(x\right)\right)^{1-\frac2\alpha}$$ Now we can iterate $$\frac{\partial^2f}{\partial x_i\partial x_j}=\frac{\partial}{\partial x_i}\left(\frac{\partial f}{\partial x_j}\right)=\frac{\partial}{\partial x_i}\left(\alpha x_j\left(f\left(x\right)\right)^{1-\frac2\alpha}\right)=\alpha\frac{\partial x_j}{\partial x_i}\left(f\left(x\right)\right)^{1-\frac2\alpha}+\alpha x_j\frac{\partial}{\partial x_i}\left(\left(f\left(x\right)\right)^{1-\frac2\alpha}\right)\\ =\alpha\left(f\left(x\right)\right)^{1-\frac2\alpha}\frac{\partial x_j}{\partial x_i}+\alpha x_j\left(1-\frac2\alpha\right)\left(f\left(x\right)\right)^{-\frac2\alpha}\frac{\partial f}{\partial x_i}\\ =\alpha\left(f\left(x\right)\right)^{1-\frac2\alpha}\frac{\partial x_j}{\partial x_i}+\alpha x_j\left(1-\frac2\alpha\right)\alpha x_i\left(f\left(x\right)\right)^{1-\frac2\alpha}\left(f\left(x\right)\right)^{-\frac2\alpha}$$ Using $\frac{\partial x_j}{\partial x_i}=\delta_{ij}$ (as it is $1$ when $i=j$ and $0$ when $i\leq j$). So we have $$ \frac{\partial^2f}{\partial x_i\partial x_j}=\alpha\left(\delta_{ij}+\left(\alpha-2\right)x_ix_j\left(f\left(x\right)\right)^{-\frac2\alpha}\right)\left(f\left(x\right)\right)^{1-\frac2\alpha} $$

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