Fundamental Theorem of Calculus real analysis Let $f$ be a continuous real-valued function on $[a,b]$ and define $H$ on $[a,b]$ by $H(x)=\int^b_xf.$ Find $H'(x)$.
I started by using the Fundamental Theorem of Calculus since this is the chapter that it comes from in the book I have. Now I'm either doing something right or horribly wrong. Any corrections or information you have on this question would be appreciated. Below is what I have attempted as of now.
Now since $f$ is continuous on the interval $[a,b]$, we know that $H(x)$ is differentiable on the interval by the FTC. Also if $f$ is continuous at a point $c \in [a,b]$, then F is differentiable at $c$ and $F'(c)=f(c)$. So based on the information given we know that $H'(x)$ does exist. 
So I then used the fact that $\int^b_af(x)=F(b)-F(a)$ and I get that $H(x)=\int^b_xf(x)=H(b)-H(x)$. I used H in this case because the book stated, If $f$ is continuous on $[a,b]$ and G is any antiderivative of $f$, then $\int^b_af=G(b)-G(a)$, so I figured that this was the correct way to interpret this question. 
From here do I really just take the derivative of $H$?
I did and got $H'(x)=\frac{d}{dx}\int^b_xf=\frac{d}{dx}[H(b)-H(x)]$. Is this completed or incorrect, I don't see where to go from here.  
 A: I suspect you are supposed to write down $$\frac{H(x)-H(x_0)}{x-x_0}$$ and to prove that the limit as $x \to x_0$ exists and is finite. It is not really difficult, you need to use the additivity of the integral with respect to the domain and the continuity of $f$ at $x_0$. You can find a proof in Rudin's book Principles of mathematical analysis, if you don't succeed.
Edit: I'll clarify my suggestion. There are two ways of solving this exercise. The first one goes as follows: let $F$ be a primitive of $f$ (on $[a,b]$), and notice that $H(x)=F(b)-F(x)$. Hence $H'(x)=0-F'(x)=-f(x)$ for every $x \in (a,b)$. In this case, the exercise is really, really trivial.
The second way assumes that you are supposed to prove that $H$ is differentiable according to the definition:
$$
H'(x)=\lim_{x \to x_0} \frac{H(x)-H(x_0)}{x-x_0} = \lim_{x \to x_0} \frac{\int_x^b f - \int_{x_0}^b f}{x-x_0} = \lim_{x \to x_0} \frac{\int_x^{x_0}f}{x-x_0}.
$$
Now you have to evaluate this limit. Actually, this is the proof of the fact that $-H$ is a primitive of $f$.
