# Infinitesimal generator as a derivation in $SO_2$

Suppose we are looking at $SO_2(\mathbb{R})$. The infitismal generator can eb found using Taylor approximation in form of a matrix, $$X=\begin{pmatrix}0&1\\-1&0\end{pmatrix}.$$ However I saw another definition for infitisimal generators by derivations and flows: Suppose $\theta_t$ is a flow which corresponds to the action of the lie group, $$X_p(f)=\lim_{\Delta t\to 0} \frac{1}{\Delta t}(f(\theta_{\Delta t}(p)-f(p)).$$ I tried to do it also here: \begin{align}X_p=\lim_{\Delta t\to 0} \frac{1}{\Delta t}(f(x \cos t+y\sin t,-x\sin t+y\cos t)-f(x,y))\\ =\lim_{\Delta t\to 0} \frac{1}{\Delta t}(f(x+y\tan t,-x\tan t+y)-f(x,y))\\ \lim_{\Delta t\to 0} \frac{1}{\Delta t}(f(x +y\tan t,-x\tan t +y)-f(x+y \tan t,y)+f(x+y\tan t,y)-f(x,y))\end{align} and here seemingly I get sum of two partial derivatives, by x and y but I don't really know how to continue.

How can I finish the proof?

Let $O_\theta$ denote the matrix representation of rotation by $\theta$. That is$$O_\theta=\left(\begin{array}{cc}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{array}\right).$$Taking the derivative at $\theta=0$ (i.e. the derivative at the identity),$$X=\left.\frac{d}{d\theta}\right|_{\theta=0}O_\theta=\left(\begin{array}{cc}-\sin0&-\cos0\\\cos0&-\sin0\end{array}\right)=\left(\begin{array}{cc}0&-1\\1&0\end{array}\right).$$