Derivative of the modulus squared of $\ f : \mathbb{R} \rightarrow \mathbb{C}$.

Question

In particular let’s say that our function $f$ described in the title is $f = f(t)$. What would be $\frac{\text{d} f}{\text{d} t}(||f(t)||^2)$?

I’ve seen that when we deal with complex functions the modulus function isn’t analytic. What about functions whose domain is $\mathbb{R}$ but whose codomain is $\mathbb{C}$?

Answer 1

I'm not completely sure what you mean, but I think what you are looking for (well, the closest you can get in general) is the following equality.

$$ \frac{\mathrm{d}}{\mathrm{d}t}(\lVert f(t)\rVert^2)=2\lVert f(t)\rVert\frac{\mathrm{d}}{\mathrm{d}t}\lVert f(t)\rVert. $$

More generally, if $g$ is an arbitrary function, then $(g^2)'=2gg'$. In this case, $g(t)=\lVert f(t)\rVert$.

Answer 2

$$ f(t) = x(t) + i y(t) \implies || f(t)||^2 = x^2(t) + y^2(t)$$

Thus $$ \frac {d}{dt} || f(t)||^2 = 2x(t) x'(t) +2y(t) y'(t)$$