describe ring of derivation of continuous map on manifold

Question

$$\\$$ Let $M$ be an arbitrary manifold and let $R$ be ring of continuous functions on it. I need to prove that $Der_{\mathbb{R}}(R, R) = 0$, (for all manifolds, and here I use Grothendieck definition of derivation, i.e. we have a field $k$, $R$ is a $k$-algebra, so derivation is $D: R \rightarrow R$ homomorphism with leibnitz) But I think I find a counterexample. $$\\$$ Let $M = \mathbb{R}$, I can prove that for all derivations $D$, $D(P(x)) = 0$, where $P(x)$ - polynomial and next I wanted to choose arbitrary function $f$ (not a polynomial one) and prove that $D(f) = 0$ (I can do it for $|x|$), but than I find 2 continuous functions ($sin(x)$ and $cos(x)$) and there is nothing I can do. $$\\$$ So I decided to make a counterexample. If I take derivation $D$ which equals zero on all functions which is not in subring generated by $sin$ and $cos$ (and on $sin$ and $cos$ work as it should be), will it work? My trouble is to check that all correct. $$\\$$ Is it correct that I should create a subring in $\mathbb{R}$ with 2 generators $sin$ and $cos$ and it will be done? Or my counterexample totally incorrect and the statement above is true? (about classification of derivation)