TITLE: Fundamental theorem of calculus with functions of $x$
QUESTION [5 upvotes]: I have these two homework questions.Using fundamental theorem of calculus and basic (A-level) facts about intergration find $F'(x)$ when:
$$F(x)=\int_{a(x)}^{b(x)} f(t) dt$$
I solved this by appling the chain rule and splitting the intergral (this might have a different name)
$$F(x)=\int_{a(x)}^{b(x)} f(t) dt=F(x)=\int_{a(x)}^{c} f(t)dt+\int_c^{b(x)} f(t) dt
=-\int_{c}^{a(x)} f(t)dt+\int_c^{b(x)} f(t) dt$$
So $$F'(x)=f(b(x))b'(x)-f(a(x))a'(x)$$ by the chain rule.
Then question 2:
$$F(x)=\left({\int_{a(x)}^{b(x)} f(t) dt}\right)^2$$
Am I correct in just saying $$F'(x)=\sqrt{f(b(x))b'(x)-f(a(x))a'(x)}$$
or is some more complex maths required? 
So my questions are: 
Is my answer for question one correct (and if not how would I go about fixing it)?
Is my answer for question 2 correct (and if not how would I go about fixing it)?
I know that homework questions are quite controversial but I have clearly pointed out this IS a homework question AND I have made an attempt at both of the questions. Any help would be greatly appreciated.

REPLY [2 votes]: Your answer to question 1 is correct. For question 2 set
$$G(x)=\int_{a(x)}^{b(x)} f(t) dt.$$
Then $F(x)=(G(x))^2$. Compute $F'(x)$ by the  chain and power rules.