Proving an integral/derivative identity.

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function, and $a<b$ real numbers. I wanted to prove that for $g:\mathbb{R}\to\mathbb{R}$ defined as $g(x)=\int^{b}_{a}f(x+t)\,dt$, $g$ is differentiable and $g^{\prime}(x)=f(x+b)-f(x+a)$ holds for all $x\in\mathbb{R}$.

This is what I've attempted: let $\varphi:[a,b]\to\mathbb{R}$, defined as $\varphi(t)=x+t$. Since $f$ is continuous, by the formula of integration by substitution for definite integrals, we have:

$$g(x)=\int^{b}_{a}f(x+t)\,dt=\int^{b}_{a}f(\varphi(t))\varphi^{\prime}(t)\,dt=\int^{\varphi(b)}_{\varphi(a)}f(u)\,du=(\ast)$$

now, by definition of definite integral,

$$(\ast)=F(\varphi(b))-F(\varphi(a))=F(x+b)-F(x+a)$$

where $F$ is any antiderivative of $f$. Now, we know that $F$ is such a function that $F^{\prime}(x)=f(x)$ for all $x$ belonging to the domain of $F$. Since $F$ is defined on the same domain as $f$, we know that $F$ is differentiable on $\mathbb{R}$. Furthermore, we know that $F(x+b)$ and $F(x+b)$ are compositions of a differentiable of function and a linear function, which is also differentiable. Hence by the chain rule we have that $F^{\prime}(x+b)=f(x+b)$, and similarly $F^{\prime}(x+a)=f(x+a)$. Thus, $g$, as a difference of two functions differentiable on $\mathbb{R}$, is also differentiable on $\mathbb{R}$, and

$$g^{\prime}(x)=F^{\prime}(x+b)-F^{\prime}(x+a)=f(x+b)-f(x+a)$$

holds for all real $x$. However, I was wondering if the above proof is complete... is there any thing missing in it? I would be thankful for some feedback.

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thus $$g'(x)=\int\limits_a^b f'(x+t).dt;$$ .
So, I think it's differentiability depends on $f'(x)$
Well, I know that the statement I wrote above (which I'm trying to prove) is true. I don't really see why differentiability of $g$ would depend on $f^{\prime}(x)$. –  Johnny Westerling Mar 20 '13 at 10:26