Find $(f^{-1})'(0)$ if $f(x) = \int_1^x{ \cos(\cos t)dt}$
For the problem there was no interval given so that the function $\cos(\cos t)$ was strictly increasing (which we need for inverse) Can I just pick any interval that works, say $0$ to $\pi/2$?
Secondly I know by FTC that since $\cos(\cos t)$ is continuous that $f$ is differentiable. How does the integral integral over $[1,x]$ come into play here? Do I need to take an antiderivative? Any hints would be wonderful.
On
Let $\;F\;$ be a primitive of $\;\cos\cos(t)\;$ , then
$$y=f(x):=\int_1^x\cos\cos t\;dt=F(x)-F(1)\implies y'=f'(x)=F'(x)=\cos\cos x\implies$$
$$\left(f^{-1}\right)' (0)\stackrel{\text{why?}}=\left.\frac1{f'(x)=0}\right|_{x\;s.t.\;f(x)}=\left.\frac1{\cos\cos x}\right|_{x\;s.t.\; f(x)=0}=\frac1{\cos\cos 1}$$
You don't want the inverse of $\cos(\cos t)$, but of $f$ defined by the integral.
Since $-1\le\cos t\le1$, $\cos u>0$ for $-\pi/2<u<\pi/2$ and $\pi/2>1$, we know that the function $t\mapsto\cos(\cos t)$ is positive (and definite for every $t$).
Therefore $f$ is increasing, because by the fundamental theorem of calculus, $$ f'(x)=\cos(\cos x)>0. $$
Now, the theorem about the derivative of the inverse function says that $$ (f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}=\frac{1}{\cos(\cos(f^{-1}(y))} $$ so all we need is to compute $f^{-1}(0)$. In other word, find $x$ such that $$ f(x)=0 $$ which shouldn't be difficult.