Find the derivative of $F^{-1}$ if $ F(x) = \int_{1}^x \frac{1}{t}dt$ 
Find the derivative of $F^{-1}$ if $\displaystyle F(x) = \int_{1}^x \dfrac{1}{t}dt.$

Attempt:
We know by the inverse theorem that $(F^{-1})'(F(x)) = \dfrac{1}{F'(x)} \implies (F^{-1})' = \dfrac{1}{F'(x)F(x)}$. Then we can proceed. My books says that $(F^{-1})'(x) = \dfrac{1}{F'(F^{-1}(x))}$. I don't see how that is the same as what I got before.
 A: The first formula you wrote: $(F^{-1})'(F(x)) = \frac{1}{F'(x)}$ is expressing $(F^{-1})'$ evaluated at $F(x)$; not $(F^{-1})'$ times $F(x)$.
Starting from $$(F^{-1})'(F(x)) = \frac{1}{F'(x)}$$ if we put $y = F(x)$, then $x = F^{-1}(y)$, so plugging this in gives $$(F^{-1})'(y) = \frac{1}{F'(F^{-1}(y))}$$ which is exactly the second equation.
A: First of all we should understand the mechanism to calculate the derivative of inverse functions. We have by definition of inverse function $$F^{-1}(F(x)) = x\tag{1}$$ for all values of $x$ in domain of $F$. Differentiating the above equation via chain rule we get $$(F^{-1})'(F(x))F'(x) = 1$$ and thus $$(F^{-1})'(F(x)) = \frac{1}{F'(x)}$$ Putting $F(x) = y$ so that $x = F^{-1}(y)$ we get $$(F^{-1})'(y) = \frac{1}{F'(F^{-1}(y)}$$ Replacing symbol $y$ by $x$ we get $$(F^{-1})'(x) = \frac{1}{F'(F^{-1}(x))}\tag{2}$$ which is what your textbook says.
For the current problem we have $F'(x) = 1/x$ and therefore from $(2)$ we get $$(F^{-1})'(x) = F^{-1}(x)$$ So the derivative of $F^{-1}$ is the function $F^{-1}$ itself.
A: $\int_1^x \frac{1}{t}dt = \ln(x) - \ln(1) =  \ln(x).$ $Re(\frac{1}{x-1}))>=0||Re(\frac{1}{x-1}))<=-1)(x\notin R||Re(1/(x-1))<-1||Re(x)>=1))||\frac{1}{x-1} \notin R $ 
So, $F(x) = \ln(x) $.
The inverse function of $\ln(x)$ is $e^x$.
$[e^x]' = e^x$
