So, in our lesson about derivatives, we learnt that $$\frac{\mathrm{d}\left|x\right|}{\mathrm{d}x}=\frac{x}{\left|x\right|}$$ This is just an application of the chain rule, since $\left|x\right|=\sqrt{x^2}$.
The above looks suspiciously similar to the formula for a unit vector, which is$$\hat{x}=\frac{\vec{x}}{\left|\vec{x}\right|}$$ and in fact, if $\vec{x}$ is a one dimensional vector (on the real number line), $x$ can be taken to be just a number, which gives us the first relation.
Is it right to reason that this can be extended to the complex numbers as well? That would mean if $$x=a+bi$$ then $$\frac{\mathrm{d}\left|x\right|}{\mathrm{d}x}=\frac{a+bi}{\sqrt{a^2+b^2}}$$ or in general, $$\frac{\mathrm{d}\left|\vec{x}\right|}{\mathrm{d}x}=\hat{x}$$
Am I right? Or is there a flaw in my reasoning? I'm asking because it's just an observation. I would be glad if someone could explain this.
While vector analysis is a bit more complicated than real analysis, there's one vector derivative that your “derivative of a scalar by a vector” could be interpreted as, and that's the gradient.
The gradient is given by the vector whose components are the partial derivatives by the coordinates $x$, $y$ and $z$ (in dimensions other than $3$, of course more or less coordinates appear). Note that for this, you can just consider the partial derivatives as derivatives where all other coordinates are treated as constants.
And we indeed have: \begin{align} \operatorname{grad}\left|\vec x\right| &= \pmatrix{\frac{\partial \left|\vec x\right|}{\partial x}\\\frac{\partial \left|\vec x\right|}{\partial y}\\\frac{\partial \left|\vec x\right|}{\partial z}} = \pmatrix{\frac{\partial\sqrt{x^2+y^2+z^2}}{\partial x}\\\frac{\partial\sqrt{x^2+y^2+z^2}}{\partial y}\\\frac{\partial\sqrt{x^2+y^2+z^2}}{\partial z}}\\ &=\pmatrix{\frac{2x}{2\sqrt{x^2+y^2+z^2}}\\\frac{2y}{2\sqrt{x^2+y^2+z^2}}\\\frac{2z}{2\sqrt{x^2+y^2+z^2}}}=\frac{\vec x}{\sqrt{x^2+y^2+z^2}}\\ &=\frac{\vec x}{\left|\vec x\right|} \end{align}
Note that in one dimension, the gradient reduces to the usual derivative, so the connection you saw is indeed not accidental.
On the second point, complex derivatives work a bit differently, and therefore you cannot get that formula. Indeed, you will just find that the complex derivative of the absolute value function does not exist.