# whats the order of a distributional derivate?

I have to calculate the derivatives of order $\le 2$ of for example $f(x) = |x|$, is it the same as the second derivate, what does this "of order $\le 2$" mean? calculating distributionell derivatives is easy, for my example it would be $$\langle f', \phi \rangle = - \int_{-\infty}^0 -1 \phi' \mathrm{d}x - \int_{0}^{\infty} 1 \phi' \mathrm{d}x = 2\int_{0}^{\infty} \phi' \mathrm{d}x = -2\phi(0)$$ is guess.

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for $f(x)=[x]$ from (-infinity,0) $f(x)=-x$ and $f'(x)=-1$,from [0,infnity) $f(x)=x$ so $f'(x)=1$ –  dato datuashvili Jun 26 '12 at 11:23
By the way, your computation of the first distributional derivative is incorrect. $$\langle f',\phi\rangle = \int_{-\infty}^0 x\phi' \mathrm{d}x - \int_{0}^\infty x\phi' \mathrm{d}x$$ Perhaps you intended to compute the second distributional derivative of $f$ in your question statement? –  Willie Wong Jun 26 '12 at 11:29

We write "of order $N$" instead of the $N$th derivative because, among other reasons, (1) that in the multivariable context there isn't just the second derivative. When your domain is $\mathbb{R}^2$ there are three independent second derivatives; (2) it makes possible for expression like "of order $\leq N$" which is a short hand for "the collection of first, second, ... , all the way up to $N$th derivatives".