Time derivative of heat semigroup. Imagine we have the heat semigroup $\{P_t\}_{0\leq t\leq T}$ and remember that we have
$$\frac{d}{dt} (P_t\varphi)(x)=\frac 1 2 (P_t\varphi '')(x).$$
I want to calculate following time derivative
$$\frac{d}{dt} P_t\varphi (\int_0^t f(u)du).$$
The problem is that the argument is also time dependent, and honestly I am having some difficulties to see why this should equal
$$\frac 1 2 (P_t\varphi '')(\int_0^t f(u)du)+P_t\varphi'(\int_0^t f(u)du)f(t)$$
This seems pretty much a product rule for derivatives, but this is not a product but the action of an operator.
I apologize if this is rather trivial but my brain seems to refuse to understand it.
I would appreciate any help.
EDIT:  Could this be due the fact that $P_t$ is a linear operator, and an application of Riesz Theorem?
 A: Approach 1: Define $$F(s,t) := (P_t \varphi) \left( \int_0^s f(r) \, dr \right).$$ Then $$\partial_t F(s,t) = (P_t \varphi'')\left(\int_0^s f(r) \, dr \right)$$ (because of the first identity in your question) and $$\partial_s F(s,t) =  \underbrace{\partial_x(P_t \varphi)(x)}_{=(P_t \varphi')(x)} \bigg|_{|x=\int_0^s f(r) \, dr} \cdot \underbrace{\partial_s \int_0^s f(r) \, dr}_{=f(s)}$$ (by chain rule). Assuming that we are dealing with sufficiently nice functions (which we need anyway, e.g. to differentiate $\varphi$), we find that the partial derivatives are continuous, and it follows that $F$ is continuously differentiable (as a function of 2 variables). Now noting that $$(P_t \varphi) \left( \int_0^t f(r) \, dr \right) = F(G(t))$$ for $G(t)=(t,t)$, an application of the standard chain rule proves the assertion.
Approach 2: By definition, $$(P_t \varphi)(x) = \frac{1}{\sqrt{2\pi t}} \int \varphi(x+y) \exp \left(- \frac{y^2}{2t} \right) \, dy,$$ and so $$(P_t \varphi) \left( \int_0^t f(r) \, dr \right) = \frac{1}{\sqrt{2\pi t}} \int \varphi \left( \int_0^t f(r) \, dr +y \right) \exp \left(- \frac{y^2}{2t} \right) \, dy.$$ In order to differentiate with respect to $t$, you need to differentiate the parameter-dependent integral. There are some general rules to do that; essentially you need a suitable upper bound for the derivative of the integrand. This approach requires some more calculations but it will, of course, lead to the same formula for the derivative.
