I am preparing for a comprehensive at the end of the month, so I would appreciate any input I could get on this solution. I am pretty confident if the first part, but I think the second answer could use some tweaks. I am not confident that I have used the assumption of absolute continuity correctly.
QUESTION: Let $f:R\to R$ be continuous and $g:[a,b]\to R$ be absolutely continuous. Show that the function $$ F(x):=\int_0^{g(x)}f(t)\,dt $$ is absolutely continuous and find $$ \frac{d}{dx}F(x) $$ for almost all $x\in[a,b]$. $$
PROOF: Since $g$ is absolute continuous on the compact set $[a,b]$, $g$ is continuous on the compact set $[a,b]$, and so we have $|g(x)|\le M_0$ for all $x\in[a,b]$ and there exists $x_1,x_2\in[a,b]$ such that $-M_0\le g(x_1)=\min_{x\in[a,b]}g(x)$ and $g(x_2)=\max_{x\in[a,b]}g(x)\le M_0$. Thus we have $f(t)$ continuous on $[g(x_1),g(x_2)]$, and so $|f(x)|\le M$ on $[g(x_1),g(x_2)]$.
Let $\varepsilon>0$ and choose $\delta>0$ such that for any $\{[a_k,b_k]\}_{k=1}^n\in[a,b]$ that satisfies $\sum_{k=1}^n|b_k-a_k|<\delta$ we have $\sum_{k=1}^n|f(b_k)-f(a_k)|<\frac{\varepsilon}{M}$. Then $$ \begin{align*} \sum_{k=1}^n|F(b_k)-F(a_k)|&=\sum_{k=1}^n\left|\int_0^{g(b_k)}f(t)\,dt-\int_0^{g(a_k)}f(t)\,dt\right|\\ &=\sum_{g(b_k)>g(a_k)}\left|\int_{g(a_k)}^{g(b_k)}f(t)\,dt\right|+\sum_{g(b_k)\le g(a_k)}\left|\int_{g(b_k)}^{g(a_k)}f(t)\,dt\right|\\ &\le\sum_{g(b_k)>g(a_k)}M|g(b_k)-g(a_k)|+\sum_{g(b_k)\le g(a_k)}M|g(a_k)-g(b_k)|\\ &=M\sum_{k=1}^n|g(b_k)-g(a_k)|\\ &<M\cdot\frac{\varepsilon}{M}\\ &=\varepsilon. \end{align*} $$ Thus $F(x)$ is absolutely continuous.
Since $F(x)$ is absolutely continuous on $[g(x_1),g(x_2)]$ we have $F(x)=\int_0^{g(x)}f(t)\,dt=f(g(x))-f(0)$, so for almost all $x\in[a,b]$ we have \begin{align*} \frac{d}{dx}F(x)&=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}\\ &=\lim_{h\to 0}\left(\frac{\int_0^{g(x+h)}f(t)\,dt-\int_0^{g(x)}f(t)\,dt}{g(x+h)-g(x)}\cdot\frac{g(x+h)-g(x)}{h}\right)\\ &=\lim_{h\to 0}\frac{\int_{g(x)}^{g(x+h)}f(t)\,dt}{g(x+h)-g(x)}\cdot\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ &=\lim_{h\to 0}\frac{f(g(x+h))-f(g(x))}{g(x+h)-g(x)}\cdot\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ &=f'(g(x))\cdot g'(x). \end{align*}
EDIT TO SECOND PART:
From part (a) we know that $F(x)$ is absolutely continuous, so for almost all $x\in[a,b]$ we have \begin{align*} \frac{d}{dx}F(x)&=\lim_{h\to 0}\frac{F(x+h)-F(x)}{h}\\ &=\lim_{h\to 0}\frac{\int_0^{g(x+h)}f(t)\,dt-\int_0^{g(x)}f(t)\,dt}{h}\\ &=\lim_{h\to 0}\frac{\int_{g(x)}^{g(x+h)}f(t)\,dt}{h}. \end{align*} Since $f(t)$ is continuous on $[g(x),g(x+h)]$ for all $h>0$, then for any $\varepsilon>0$ and $h$ small enough, we have $|f(t)-f(g(x))|<\varepsilon$. So \begin{align*} \lim_{h\to 0}\frac{\int_{g(x)}^{g(x+h)}f(t)\,dt}{h}&=\lim_{h\to 0}\frac{f(g(x))[g(x+h)-g(x)]}{h}\\ &=f(g(x))\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\\ &=f(g(x))\cdot g'(x). \end{align*}
Formalizing the continuity argument is escaping me at the moment...
Calculation of $F'(x):$
Note that since $f$ and $g$ are continuous, $f(t)$ for $g(x) \leq t \leq g(x+h)$ is "almost equal to" $f(g(x))$ for small $h$.
(You can make this precise by using epsilons and deltas, but I think the intuition is really important here).
So, $F'(x)= \lim_{h \rightarrow 0}(f(g(x))).(g(x+h)-g(x))/h$. Since $g$ is absolutely cont. $\lim_{h \rightarrow 0}(g(x+h)-g(x))/h = g'(x)$ a.e., and hence $F'(x)=f(g(x)).g'(x)$ a.e.