In Lemmermeyer's class field theory notes, on page 114 there is the following claim. Let $F = \mathbb Q(\sqrt{-5})$, then let $K = F(\sqrt{-1})$. Let $F_1 = \mathbb Q(\sqrt{-1})$ and $F_2 = \mathbb Q(\sqrt{5})$ be the other two quadratic subfields of $K/\mathbb Q$. Then we have that $$ \operatorname{diff}(K/F) \mid \operatorname{diff}(F_1/\mathbb Q) \quad\text{and}\quad \operatorname{diff}(K/F) \mid \operatorname{diff}(F_2/\mathbb Q). $$
Why is this true? Certainly I see $\operatorname{diff}(K/F)$ divides $\operatorname{diff}(K/\mathbb Q) = \operatorname{diff}(K/F_i)\operatorname{diff}(F_i/\mathbb Q)$ for $i=1,2$ but I don't see how to get the claim.
The point is that $K/F = F(\sqrt{-1})$ is a "relativization" of $F_1/\mathbb{Q} = \mathbb{Q}(\sqrt{-1})$, so in particular, (i) any element generating the latter also generates the former; (ii) the trace form on the latter extends to the trace on the former. (For $F_2/\mathbb{Q}$ it's the same.)
Now to show the different of the former divides the different of the latter, we can either use (i) together with the "$(f'(\alpha))$ definition" of different, or (ii) together with the "inverse of inverse different" definition (see Wikipedia, especially the Neukirch reference given (p. 198, Chapter III, Theorem 2.5) for the equivalence of definitions).
(For the special case that "relativization" preserves unramified-ness, see Neukirch p. 153, Ch. II, Sec. 7, Proposition 7.2.)