Diffrentiate or integrate? I'm so confused

We have

$\displaystyle e^{-x}f(x) = 2 + \int\limits_0^x \sqrt{t^4 + 1} dt$

At $x=0 , f(x)=2$ let $u=g(x)$ be the inverse of $y=f(x)$ . Therefore, $fog = x$ or (\f(u) = x )

which implies that, at $x=2 , u=g(2)=0$ .

In the given equation, replace $x$ with $u$ . Thus, $\displaystyle e^{-u}f(u) = 2 + \int\limits_0^u \sqrt{t^4 + 1} dt$

$\displaystyle e^{-u}x = 2 + \int\limits_0^u \sqrt{t^4 + 1} dt$

Using lebnitz theorem, differentiating this equation, we get,

$\displaystyle e^{-u}(1-\frac{xdu}{dx}) = \frac{du}{dx}\sqrt{u^4 + 1}$

we have to evaluate derivative of inverse of $f(x)$ , i.e. , $\frac{du}{dx}$

$\displaystyle \frac{du}{dx} = \frac{1}{e^u\sqrt{u^4 + 1} + 2}$

$\displaystyle (\frac{du}{dx})_{x=2} = \frac{1}{1\sqrt{0 + 1} + 2}$

$\displaystyle (\frac{du}{dx})_{x=2} = \frac{1}{3}$

$\displaystyle (f^{-1})'(2) = \frac{1}{3}$

Thus, $\boxed {3(f^{-1})'(2) = 1}$