$g(x)$ is a differentiable function everywhere but at a point $x_{0}$ where it has a discontinuity with a jump:
$\Delta g_{0} = \lim_{\epsilon \to 0} [g(x_{0}+ \epsilon)-g(x_{0}- \epsilon)]$
How to show that the derivative of $g_{x}$ as a distribution is
$g'(x) = g'(x)_{cl} + \Delta g_{0}\delta(x-x_{0})$
where $g'(x)_{cl}$ is the usual derivative for $g(x)$ for $x \neq x_{0}$
Consider the function $H(x) = \Delta g_0 \mathcal I_{x_0>0}$ where $\mathcal I$ is an indicator function. We show that $H'(x) = \Delta g_0 \delta(x-x_0)$ in the distributional sense. Consider the action on a test function $\varphi \in C_{supp}^\infty$: \begin{align} \int_{-\infty}^{\infty} H'(x) \varphi(x) \,dx & = \underbrace{H(x) \varphi(x) \big|_{x =-\infty}^{\infty}}_{= 0 \text{ by compact support}} - \int_{-\infty}^{\infty} H(x) \varphi'(x) \, dx \\ & = - \int_{x_0}^{\infty} \Delta g_0 \varphi'(x) \, dx \\ & = \Delta g_0 \varphi(x_0) \\ & = \int_{-\infty}^{\infty} \Delta g_0 \delta(x - x_0) \varphi(x) \, dx. \end{align} This proves the above statement, your conclusion now follows easily by adding a differentiable function to $H(x)$ (except maybe at $x = x_0$).