Prove: using the Fundamental theorem of calculus

Question

I have trouble doing proofs using the fundamental theorem of calculus and I think seeing an example would help.

Suppose we have a continuous function $ f $ defined on a real interval and a function $ F $ that is also defined on the same interval; and $F(x)=\int^b_x fdt$. Then for $ x $ in that interval, $ F'(x)=-f(x) $.

I know this can be done using the fundamental theorem of calculus; how is it applied?

By the way, the theorem says this:

Let $ f $ be integrable on $[a,b]$. For $x\in[a,b]$ put $F(x)=\int^x_a f(t)dt$. Then $ F$ is continuous on $[a,b]$ and furthermore if $ f $ is continuous at a point $x_0$ of $[a,b]$ then $F$ is differentiable at $x_0$ and $F'(x_0)=f(x_0)$.

Answer 1

We name the function from FTC $F$ and our new function $G$ then $$G(x) = \int_x^b f(t)\ \mathrm dt = \underbrace{\int_a^b f(t)\ \mathrm dt}_{\text{constant}} - \int_a^x f(t) \ \mathrm dt = C - F(x)$$ So $G'(x) = -F'(x) = -f(x)$ by FTC.

Answer 2

if F(x)=∫bxfdt, then -F(x)=∫xbfdt. by the FTC if -F(x)=∫xbfdt, then (-F(x))'=f(x), but (-F(x)')=-F'(x), so F'(x)=-f(x).