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Define a function $f(x)$ on $\mathbb{R}$ as:

$f(x) =\begin{cases}0&\text{for }x \le 0\\\\\frac{1}{\sqrt{x}}&\text{for }x > 0\;.\end{cases}$

Then, how can I calculate the weak derivative(of course, distribution sense) of it? I roughly guess in $x>0$ region it becomes $\dfrac{-1}{2(\sqrt{x})^3}$; but I don't know how to deal with the neighborhood of $x=0$.

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Since $f$ is integrable on $\mathbb{R}$, by definition, it induces a distribution $T_f$, such that for any test function $\phi$ $T_f(\phi)=\int_{-\infty}^{+\infty}f\phi dx\quad and\quad T'_f(\phi)=-\int_{-\infty}^{+\infty}f\phi'dx,$ where $T_f'$ is the distributional derivative of $f$. Note that $f'(\phi-\phi(0))$ is integrable on $\mathbb{R}$, so integration by parts implies that $T'_f(\phi)=\int_{-\infty}^{+\infty}f'(\phi-\phi(0))dx.$