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Let us define $g: [-1,1] \rightarrow \mathbb{R}$ $$g(x) = \left\{ \begin{array}{l l} \frac{1}{\sqrt{x}} & \quad \text{if } x\in (0,1]\\ 0 & \quad \text{if } x=0\\ 1-x^2-x & \quad \text{if } x\in [-1,0)\\ \end{array} \right. \\$$ I would like to find the distributional derivative of $g$. After seeing the standard examples $|x|$ and Heaviside function. I made up this function to test myself, but I am stuck... Here is what I have so far.

Note that since $g\in L^1_{loc}([-1,1])$, the distributional derivative of $T_g$ can be explicitly written as $$\langle T_g ', \phi \rangle := -\langle T_g, \phi'\rangle = -\int_{-1}^1 g\phi' dx.$$ By splitting up the integral to $-\int_{-1}^0$ and $-\int_0^1$ (here we ignore that $g(0) = 0$), we compute the following $$- \int_{-1}^0 (1-x^2 -x) \phi' dx = - (1-x^2-x)\phi\bigg|_{-1}^0 + \int_{-1}^0 (-2x-1) \phi dx = -\phi(0) + \int_{-1}^0 (-2x-1) \phi dx.$$ Now however, since $\frac{1}{\sqrt{x}}$ is not absolute continuous in $(0,1]$, the integration formula fails here, $\Big(\frac{1}{\sqrt x}\Big)' = \frac{-1}{2x^{3/2}}$ is not integrable. $$-\int_0^1 \frac{1}{\sqrt{x}} \phi' dx = - \bigg(\frac{1}{\sqrt{x}}\bigg)\phi\bigg|_0^1 + \int_0^1 \frac{-1}{2x^{3/2}}\phi dx $$ Is there anything we could do for this computation?

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$ -\int_0^1 \frac{1}{\sqrt{x}} \phi' dx = - \bigg(\frac{1}{\sqrt{x}}\bigg) (\phi - \phi(0))\bigg|_0^1 + \int_0^1 \frac{-1}{2x^{3/2}}(\phi(x) - \phi(0)) dx =\\ =-(\phi(1) - \phi(0)) + \int_0^1 \frac{-1}{2x^{3/2}}(\phi(x) - \phi(0)) dx $

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