It depends. Your function $f : \mathbb{C} \rightarrow \mathbb{C}$ can always be seen as a function from $\mathbb{R}^2$ to $\mathbb{R}^2$. If it is at least differentiable in the real sense, then it certainly makes sense to compute the derivative of $f$ as a real function, which will give you a $2x2$ real matrix.
The relation this derivative has with the complex derivative can be expressed in many ways, such as the following: you can define the operators $\frac{\partial}{\partial{}z} = \frac{1}{2}\left(\frac{\partial}{\partial{x}} -i\frac{\partial}{\partial{y}}\right),~~\frac{\partial}{\partial{}\bar{z}} = \frac{1}{2}\left(\frac{\partial}{\partial{x}} +i\frac{\partial}{\partial{y}}\right),$ which can be applied to any function $f$ having at least partial derivatives. Then, you can check that such a function $f$ satisfies the Cauchy-Riemann equations if and only if $\frac{\partial{f}}{\partial{\bar{z}}} = \frac{1}{2}\left(\frac{\partial{f}}{\partial{x}} +i\frac{\partial{f}}{\partial{y}}\right) = 0.$ You can also check that the matrix of $\mathrm{df}$ as a map $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a conformal linear mapping at every point if and only if $f$ satisfies Cauchy-Riemann. A conformal linear mapping is a linear mapping that preserves angles, intuitively speaking. Formally, $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is conformal if and only if it has the form $\alpha{}L$, where $L : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is an orthogonal linear map (that is, $L^tL = 1$).