Let $X$ be a topological space (assume additional assumptions if needed) and denote by $\mathcal O \_X$ its sheaf of $\Bbbk$-valued continuous functions where $\Bbbk$ is $\mathbb{R}$ or $\mathbb{C}$ with standard topology.


Then, as it is done in the differentiable setting or in algebraic geometry, one can define the following objects
$$T\_X:=\mathscr{Der}\_\Bbbk (\mathcal O\_X,\mathcal O\_X)$$
the tangent sheaf, i.e. the sheaf of $\Bbbk$-linear derivations of $\mathcal O\_X$ with values in $\mathcal O\_X$ (on local sections, $\Bbbk$-linear maps $D:\mathcal O\_X(U)\to\mathcal O\_X(U)$ satisfying Leibniz: $D(f\cdot g)=f\cdot Dg + g\cdot Df$), and 
$$\Omega\_X^1:=\mathcal I/\mathcal I^2$$


the sheaf of differentials, where $\mathcal I$ is the ideal sheaf of $X$ embedded diagonally $\Delta:X\hookrightarrow X\times X$ into $X\times X$ (i.e. $\mathcal I(U)=$ functions in $\mathcal O\_{X\times X}(U)$ that are zero on every point of $\Delta(X)\subset X\times X$).


Well, what can be said about these two sheaves? Anything interesting at all?


Also, is there any relationship between $T\_X$ and the "tangent microbundle" $\tau\_X$ in case $X$ is a topological manifold?



The (co)tangent sheaf of a topological space
Your $$ is always $0$. If $$ is a derivation and $$ is a function, then for every point $$ $$ vanishes at $$; it suffices to prove this when $()=0$, and in that case $=ℎ$ where $()=0=ℎ()$, so $=ℎ+ℎ$ vanishes at $$. 


