Two answers: (1) Distribution theory. On the space $\mathcal D'(\mathbb R)$ of continuous linear forms on $\mathcal D(\mathbb R)=C_c^\infty(\mathbb R)$ it is easy to define the first derivative: $$ \langle\frac{du}{dx},\phi\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}= -\langle u,\frac{d\phi}{dx}\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}. $$$$ \left\langle\frac{du}{dx},\phi\right\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}= -\left\langle u,\frac{d\phi}{dx}\right\rangle_{\mathcal D'(\mathbb R),\mathcal D(\mathbb R)}. $$ You get the ordinary derivative of a differentiable function, also $H'=\delta$ ($H$ is the Heaviside function, characteristic function of $\mathbb R_+$, $\delta$ the Dirac mass), $$ \frac{d}{dx}(\ln \vert x\vert)=\text{pv}\frac{1}{x} $$$$ \frac{d}{dx}(\ln \vert x\vert)=\operatorname{pv}\frac{1}{x} $$ and many other classical formulas. In particular, you can define the derivative of any $L^1_{loc} $$L^1_\text{loc} $ function, of course not pointwise but as above.
(2) Operator theory. In $L^2(\mathbb R)$, you consider the subspace $H^1(\mathbb R)=${$u\in L^2(\mathbb R), u'\in L^2(\mathbb R)$}$H^1(\mathbb R)=\{u\in L^2(\mathbb R), u'\in L^2(\mathbb R)\}$, where the derivative is taken in the distribution sense. Then the operator $d/dx$ is an unbounded operator with domain $H^1(\mathbb R)$. It is even possible to prove that the operator $\frac{d}{idx}$$\frac{d}{i\,dx}$ is selfadjoint.
