TITLE: Time derivative of a pullback of a time-dependent 2-form
QUESTION [8 upvotes]: I've been reading McDuff's Introduction to symplectic topology book and I couldn't figure out how to justify the last line in this image(p108):

Specifically, the identity
$$\frac{d}{dt}\psi^*_t \omega_t = \psi^*_t (\frac{d}{dt}\omega_t + \iota(X_t)d\omega_t + d\iota(X_t)\omega_t)$$
When $\omega_t = \omega$ is constant with respect to $t$, this reduces to the well-known Cartan's formula:
$$\frac{d}{dt}\psi^*_t \omega = \psi^*_t (\iota(X_t)d\omega + d\iota(X_t)\omega)$$
My question is this: how would one derive/justify the generalization?

REPLY [8 votes]: This is a frequently-used trick in differential geometry. The key idea is the following fact, taken from basic multi-variable calculus:
Let $F:\mathbb{R}^2\to V$ be a smooth function, where $V$ is some vector space, and let $dF_p$ denote the differential of $F$ at $p$. Then at every $p\in\mathbb{R}^2$ the differential $dF_p$ is linear, and in particular, we have $$dF_p(1,1)=dF_p(1,0)+dF_p(0,1).$$
How is this related to the question at hand?
We wish to calculate the time derivative $\frac{d}{dt}\psi_t^*\omega_t$. The difficulty lies in the fact that $t$ appears twice in the expression we are trying to differentiate. So, the trick is to consider the two-parameter family $$F(t,s):=\psi_t^*\omega_s,$$ where $t$ and $s$ may vary independently. Then, by the above basic fact, we have $$\frac{d}{dt}\psi_t^*\omega_t=dF(1,1)=\frac{\partial F}{\partial t}+\frac{\partial F}{\partial s},$$ which gives the desired expression.