In every complex analysis book I have come across (mainly Ahlfors and Stein/Shakarchi), the calculation $\frac{d}{dt} e^{it} = ie^{it}$ is taken for granted. But I want to look at this more carefully, because it doesn't seem immediately obvious to me.
The function $f(t) = e^{it}$ is a function $\mathbb{R} \to \mathbb{C}$, and for any function $\mathbb{R} \to \mathbb{C}$ we define the derivative by dealing with the real and imaginary parts separately: $\frac{d}{dt}(u(t) + iv(t)) := u'(t) + iv'(t)$. So working from the definition, I would compute $\frac{d}{dt}e^{it}$ by writing $e^{it} = \cos(t) + i\sin(t)$ and differentiating each component. This does give us $ie^{it}$, but what the authors of these books seem to suggest without proof is that there is a type of chain rule that can be applied to make the calculation simpler. This chain rule would have to apply to compositions $\mathbb{R} \to \mathbb{C} \to \mathbb{C}$, but the only chain rules I have seen explicitly described are for compositions of differentiable functions $\mathbb{R} \to \mathbb{R} \to \mathbb{R}$ and for holomorphic functions $\mathbb{C} \to \mathbb{C} \to \mathbb{C}$. In other places in the books, a chain rule for compositions $\mathbb{R} \to \mathbb{R} \to \mathbb{C}$ was used, but the proof of this one is trivial in light of the definition of derivatives of functions $\mathbb{R} \to \mathbb{C}$ and the ordinary real chain rule. But the case $\mathbb{R} \to \mathbb{C} \to \mathbb{C}$ is different from all of these, and I'm surprised none of the authors have mentioned it, as it doesn't seem to follow immediately from either the $\mathbb{R} \to \mathbb{R} \to \mathbb{R}$ or $\mathbb{C} \to \mathbb{C} \to \mathbb{C}$ chain rule.
Does such a chain rule actually exist? And if so, how do we prove it?
(Expanding out my comments into an answer)
First of all, I'm not sure that I agree with your definition of the complex derivative via $\frac{d}{dt}\left(u(t)+iv(t)\right)=\frac{du(t)}{dt}+i\frac{dv(t)}{dt}$. While that's true, I see it as a consequence of the canonical definition of the derivative of a function from $\mathbb{R}$ to any vector space $V$ over it, $\frac{d}{dt}f(t)|_{t_0}$ $= \lim\limits_{t\to t_0}\dfrac{f(t)-f(t_0)}{t-t_0}$. I feel like this definition is better both because it's more 'holistic' (there's no breaking the function into real and imaginary parts) and more general (it applies over more than just $\mathbb{C}$). The rule $(u+iv)'=u'+iv'$ then becomes a handy computational shortcut that falls out readily from this definition.
In the particular case $V=\mathbb{C}$, since $\mathbb{R}$ sits inside $\mathbb{C}$, we can say a lot more: the functions from $\mathbb{R}\to\mathbb{C}$ are a subset of the functions for $\mathbb{C}\to\mathbb{C}$, and in fact every sufficiently 'nice' function (specifically, every analytic function) $\mathbb{R}\to\mathbb{C}$ extends to a nice-enough function $\mathbb{C}\to\mathbb{C}$. (I'm being a little bit loose here; in particular, the function may have singularities and so not necessarily be defined on all of $\mathbb{C}$, e.g. $f(t)=\frac{1}{t^2+1}$. But it will be well-defined on an open set containing the real line.) In particular, the derivative of a function $f$ viewed as an analytic function $\mathbb{R}\to\mathbb{C}$ can be seen as the derivative of $f$ viewed as an analytic function $\mathbb{C}\to\mathbb{C}$, which is then restricted to $\mathbb{R}$. These derivatives are guaranteed to coincide, because the limit in the definition of derivative must agree over any sequence converging to $t_0$, and in particular will agree on all the sequences $t\to t_0$ where all the $t$ are in $\mathbb{R}$. So by and large, you can really view the $\mathbb{R}\to\mathbb{C}\to\mathbb{C}$ case as a special case of $\mathbb{C}\to\mathbb{C}\to\mathbb{C}$. If you're interested in non-analytic $f: \mathbb{R}\to\mathbb{C}$ then certainly things are going to get a little more complicated, but I'd say that the huge majority of functions you hit in low-level analysis texts are going to be analytic.