derivative of an integral from 0 to x when x is negative? Given a function $$F(x) = \int_0^x \frac{t + 8}{t^3 - 9}dt,$$ is $F'(x)$ different when $x<0$, when $x=0$ and when $x>0$? 
When $x<0$, is $$F'(x) = - \frac{x + 8}{x^3 - 9}$$ ... since you can't evaluate an integral going from a smaller number to a bigger number? That's what I initially thought, but when I graphed the antiderivative the intervals of increase and decrease were different from the ones in my calculation.
EDIT: Yes, I was talking about that identity. 
So is $$F'(x) = \frac{x + 8}{x^3 - 9}$$ or $$F'(x) = -\frac{x + 8}{x^3 - 9}$$ ??
EDIT: thanks for the answers! but what happens when x is bigger than 9^1/3? The derivative is also the same? Btw, when x = 0 shouldn't F(x) = 0 and hence F'(x) = 0?

No, we have
$$F'(x)=\frac{x+8}{x^3-9}$$
for all $x<\sqrt[3]{9}$. The limitation is due to the fact that the integral is meaningful only when the interval doesn't contain $\sqrt[3]{9}$ and so we must consider only the interval $(-\infty,\sqrt[3]{9})$ that contains $0$.
If $b<a$, one sets, by definition,
$$
\int_a^b f(t)\,dt=-\int_b^a f(t)\,dt
$$
so the equality
$$
\int_a^c f(t)\,dt=\int_a^b f(t)\,dt+\int_b^c f(t)\,dt
$$
without restrictions on the limits of integration, provided we don't jump over points where $f$ is not defined so that all the integrals make sense.
This relation is what the fundamental theorem of calculus relies on. Remember that, for continuous $f$, there exists $\xi$ such that
$$
\frac{1}{b-a}\int_a^b f(t)\,dt=f(\xi)
$$
where $\xi\in[a,b]$ if $a<b$. Therefore, for $h>0$,
$$
\int_0^{x+h}f(t)\,dt-\int_0^x f(t)\,dt=
\int_x^0 f(t)\,dt+\int_0^{x+h}f(t)\,dt-\int_0^x f(t)\,dt=
\int_{x}^{x+h}f(t)\,dt=hf(\xi)
$$
for some $\xi\in[x,x+h]$, so that
$$
\lim_{h\to0+}\frac{F(x+h)-F(x)}{h}=f(x).
$$
Similarly for the limit from the left.
