TITLE: Value of the Inverse Function
QUESTION [0 upvotes]: Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that:
$e^{-x}\cdot f(x)= 2 + \int_{0}^{x} \sqrt{t^4+1}dt$ ...[for $x\in (-1,1)$]
Let $f^{-1}$ be the inverse of $f$. What is the value of $(f^{-1})'(2)$? 
Differentiating the initial equation, you get $f'(x)=f(x)+e^x \sqrt{x^4+1}$.
We know that $\frac{dx}{dy}=\frac{1}{dy/dx}$, so $[f^{-1}(x)]' = \frac{1}{f(x)+e^x \sqrt{x^4+1}}$. Simply plugging $x=2$ in the equation does not help as $f(2)$ is not calculable because of the integrand is not calculable (using pen and paper). How to go about this? 
Another doubt I have is, when is it valid to simply plug in values of $x$ in derivative of the $f$ inverse equation? Because considering a simply function as $e^x$ and its inverse $\ln x$, we can clearly see plugging values of x is not valid. Is the differential of inverse function is reciprocal of differential of orginal function an incomplete statement?

REPLY [0 votes]: You are on the right path.  However, 
$[f^{-1}]'(a) = 1/f'(f^{-1}(a))$
$f^{-1}(2) = 0$
You don't need to plug 2 into, to what you have worked out so far.  You need to plug 0.
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