Given $f(x)=\int_5^x \sqrt{1+t^2}\,dt$, find $(f^{-1})'(0)$ 
If $f(x)=\int_5^x \sqrt{1+t^2}\,dt$, find $(f^{-1})'(0)$.

Here is what I have done so far. I have took $f'(x)=(1+x^2)^{1/2}$ and I have found $1/f'(0)$ which should equal $1$. I don't think this is the final answer though. I'm having trouble continuing and getting to the end.
i edited the integration for 0 because it was suppose to be 5. sorry for that
 A: Let we do this step by step.
Step 1) $\sqrt{1+t^2}$ is a continuous positive function, hence $f(x)$ is a differentiable increasing function;
Step 2) $f(5)=0$; due to Step1, the inverse function of $f$ exists in a neighbourhood of zero and it is an increasing differentiable function;
Step 3) Let $g$ be the inverse function of $f$; from Step 2 we have $g(0)=5$. Since in a neighbourhood of zero
$$ f(g(x)) = x $$
holds, differentiation gives:
$$ g'(x)\cdot f'(g(x)) = 1$$
from which:
$$ g'(0) = \frac{1}{f'(5)} $$
and we just need to find $f'(5)$.
Step 4) Let $F(t)$ be a primitive of $\sqrt{1+t^2}$. Then:
$$ f(x) = F(x) - F(5), $$
hence differentiation and the fundamental theorem of calculus give:
$$ f'(x) = F'(x) = \sqrt{1+x^2}, $$
so $f'(5)=\sqrt{26}$ and 
$$ g'(0) = \frac{1}{\sqrt{26}}.$$
A: You need $f^{-1}(0)$, which means you need $x$ for which $f(x)=0$.
Then your answer is $1/f\,'(f^{-1}(0))$
A: Hint: Substituting $$u=\sqrt{1+t^2}-t,\implies t=\frac{1-u^2}{2u}$$,
the integral becomes,
$$\begin{align}
f(x)
&=\int_{5}^{x}\sqrt{1+t^2}\,\mathrm{d}t\\
&=-\int_{\sqrt{26}}^{\sqrt{1+x^2}-x}\frac{u+\frac{1}{u}}{2}\cdot\frac{u^2+1}{2u^2}\,\mathrm{d}u\\
&=-\int_{\sqrt{26}}^{\sqrt{1+x^2}-x}\frac{(u^2+1)^2}{4u^3}\,\mathrm{d}u.
\end{align}$$
Now you have $f(x)$ written as an integral of a rational function, which can be evaluated in the usual manner. Obtain an explicit formula for $f$, and then find its inverse.
A: First notice this the function $f(x)$ is increasing for $x\in(-\infty,\infty)$ (since $f'(x)>0$ ) which implies that the inverse exists. Let's write

$$f(x) =g(x),  $$

where 

$$  g(x) = \int_{5}^{x}\sqrt{1+t^2}dt. $$

Proceeding as 
$$ f(x)=g(x) \implies x = f^{-1}(g(x)) \implies 1= (f^{-1}(g(x)))'g'(x)$$
$$ \implies (f^{-1}(g(x)))' = \frac{1}{g'(x)} . $$
Note that we want the derivative of inverse function at $0$ that means we should have $g(x)=0$ which implies $x=5$ (Check the definition of $g(x)$). So we have  

$$ (f^{-1})'  = \frac{1}{g'(0)} = \frac{1}{\sqrt{26}}. $$

A: If $(x, y)\in D_f$ then $(y, x)\in D_{f^{-1}}$. Therefore $${(f^{-1})}'(0)=\frac{1}{{(f)}'(5)}$$ also $f'(x)=\sqrt{1+x^2}$, hence $${(f^{-1})}'(0)=\frac{1}{{(f)}'(5)}\frac{1}{\sqrt{26}}$$
A: Since $f(5)=\int_5^5 \sqrt{1+t^2}\,dt=0$, and since $f(x)$ is always positive for all other $x$ since the integrand is non-negative, we uniquely have $f^{-1}(0)=5$. Now we have to determine $(f^{-1})'(0)$. We have
$f^{-1}(x)=y$
$x=f(y)$
$1=f'(y) y'$ by differentiating
$y'=\frac1{f'(y)}$
$(f^{-1})'(x)=\frac1{f'(f^{-1}(x))}$
Hence
$(f^{-1})'(0)=\frac1{f'(f^{-1}(0))}=\frac1{f'(5)}=\frac1{\sqrt{1+5^2}}=\frac1{\sqrt{26}}$ 
since $f'(x)=\sqrt{1+x^2}$ by the Fundamental Theorem of Calculus.
