Let $f$ be continuous on $\mathbb{R}$.
Consider the following sequence of functions:
$F_1(x) = \int_0^x xf(t)\operatorname dt$
$F_2(x) = \int_0^{\int_0^x xf(t)\operatorname dt} xf(t)\operatorname dt$
$F_3(x) = \int_0^{\int_0^{\int_0^x xf(t)\operatorname dt} xf(t)\operatorname dt} xf(t)\operatorname dt$
Given $\int_0^1 f(t)\operatorname dt=1$ and $f(1) = 2$.
Evaluate $F_n^{'}(1)$ for each $n ∈ \mathbb{N}.$
Make sure to justify your answer and only use FTC I in its most basic form to evaluate the derivative of integrals.
What I tried so far:
It's not hard to see that $$F_n(x)=\int_0^{F_{n-1}(x)} xf(t)\operatorname dt \text{, where } n>1$$
Let $G(x)$ be a anitiderivative of $xf(t)$
By FTC II we have $$F_n(x)=G(F_{n-1}(x))-G(0)$$
$$F_n^{'}(x)=G^{'}(F_{n-1}(x))(F_{n-1}^{'}(x))$$
$$F_n^{'}(1)=G^{'}(F_{n-1}(1))(F_{n-1}^{'}(1))$$
$$F_{n-1}^{'}(1)=G^{'}(F_{n-2}(1))(F_{n-2}^{'}(1))$$
$$\vdots$$
$$F_{2}^{'}(1)=G^{'}(F_{1}(1))(F_{1}^{'}(1))$$
$$F_1^{'}(1) = G^{'}(1)$$
And what should I do for $\int_0^1 f(t)\operatorname dt=1$ and $f(1) = 2$
How to do this "only use FTC I in its most basic form to evaluate the derivative of integrals."
Any help or hint or suggestion would be appreciated.
Hint: If $$F(x)=\int_0^{g(x)}f(t) \,dt$$then the chain rule and FTC I show that $$F'(x)=g'(x)f(g(x)).$$