I want to solve the following exercise.
Let A be an $\mathbb{R}$-algebra. A $derivation$ on A is a $\mathbb{R}$-linear map $D: A \to A$ obeying the Leibniz rule $ D(ab) = D(a)b + aD(b) $ for all $a,b \in A$.
Now let M be a manifold. Show that the algebra $ C(M) = \{ f: M \to \mathbb{R} : f \textrm{ continuous } \} $ has no non-trivial derivations, i.e. for every linear map $D : C(M) \to C(M)$ obeying the Leibniz rule, it follows that $D = 0$.
Hint: Use that every $f \ge 0$ can be written as a square.
I have some difficulties in understanding. I am currently reading about manifolds and tangent spaces, and I read that the tangent space could be defined as the space of all point derivations $D : C^{\infty}(M) \to \mathbb{R}$, but this space is more than just the trivial derivation $D = 0$, does it make a huge difference if I am considering smooth vs. just continuous functions? And in the exercise, how does the hint help me? I can write every $f \ge 0$ as $(\sqrt{f})^2$, and then $ D(g^2) = 2gD(g) = gD(g+g) $ with $g = \sqrt{f}$, does it follow that $D = 0$?.