Derivative of Integral With Limits I have the following problem:

Let $f(x) = \int_{3}^{x}\sqrt{1+t^3}\ dt$. Find $(f^{-1})'(0)$.

I know that if $g' = f^{-1}(x)$ that $g'(f(x))f'(x) = 1$. Obviously when $x$ is $3$, $f(x)$ is $0$. So I worked out that:
$\displaystyle\ g’(f(3))f’(3)=1→ g'(0)=\frac{1}{f'(3)}→(f^{-1})'(0)=\frac{1}{\frac{d}{dt}[\int\sqrt{1+3^{3}}dt]}=\frac{1}{\sqrt{1+27}}$ $$=\frac{1}{\sqrt{28}}.$$
My question is: since $f(x)$ was originally an integral evaluated from $3$ to $x$, shouldn't $x$ be $3$ across the equation, making the attempt to take the derivative of it useless? Would this result in the solution being "not defined" since you can't divide $(1)$ by zero? Or, does the derivative effectively nullify the limits?
 A: By the Fundamental Theorem of Calculus, you have $$f'(x)=\sqrt{1+x^3}.$$
Alternatively recall that $x=f(f^{-1}(x))$, so differentiating both sides of the equation and using the chain rule we have $$1=f'(f^{-1}(x))(f^{-1}(x))'$$
so $$(f^{-1}(x))'=\frac{1}{f'(f^{-1}(x))}=\frac{1}{\sqrt{1+(f^{-1}(x))^3}}$$
then since $f(3)=0$ we have $f^{-1}(0)=3$ (as you mentioned) and thus
$$(f^{-1}(0))'=\frac{1}{\sqrt{1+(f^{-1}(0))^3}}=\frac{1}{\sqrt{1+27}}=\frac{1}{\sqrt{28}}.$$
A: My whole answer here is going to be based on the fact that you've assumed $x \geq 3$. If i'm wrong your doubt should rather be related the the sense of the limits in integration.
I want you to recall the fundamental principle of derivatives.

The first principle of differentiation.

$$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{(x+h)-(x)}$$
Here we get the derivative value by running $h$ in both positive and negative neighbourhood of $0$.
Hence getting infinitesimal change in $f(x)$ in neighbourhood of $x$ over infinitesimal change in $x$.
But when either of the neighbourhood is not present in the domain near $x$ we only in one direction, instead.

For example $\ln(x), x^{\frac{3}{2}} \sqrt{x}$ etc...

You get the picture i guess. Now you'd question the sense of this derivative. It is when what will be the slope (rate of change) of the curve/function when it will go in the only direction allowed.
Similarly, here the function in question has such property.
More precise solution for such derivatives by first principle can be given as:
$$f(x)=\int_{3}^{x}{\sqrt{1+t^3}\ dt}$$
For this particular function the meaning of derivative at $x=3$ shall be:
$$f'(3)=\lim_{h \to 0^+}\frac{f(3+h)-f(3)}{h}$$

Above equation says that $x$ will only vary in positive neighbourhood of $3$.

