Let $f$ be a real-valued function defined on the interval $(-1,1)$ such that $e^{-x}f(x)=2+\int_{0}^{x}\sqrt{t^{4}+1}\,dt$, for all $x\in(-1,1)$, and let $f^{-1}$ be the inverse function of $f$. Then $(f^{-1})'(2)$ is equal to

$e^{-x}f(x)=2+\displaystyle\int_{0}^{x}\sqrt{t^{4}+1}\,dt,\quad x\in(-1,1)$

$f^{-1}:\ \text{inverse of } f$

$\text{Find } (f^{-1})'(2)$

$x=0:\ e^{0}f(0)=2+0 \Rightarrow f(0)=2$

$\Rightarrow f^{-1}(2)=0$

$(f^{-1})'(y)=\dfrac{1}{f\,'(f^{-1}(y))}$

$\Rightarrow (f^{-1})'(2)=\dfrac{1}{f\,'(0)}$

$\dfrac{d}{dx}:\ -e^{-x}f(x)+e^{-x}f\,'(x)=\sqrt{x^{4}+1}$

$\times e^{x}:\ -f(x)+f\,'(x)=e^{x}\sqrt{x^{4}+1}$

$\Rightarrow f\,'(x)=f(x)+e^{x}\sqrt{x^{4}+1}$

$f\,'(0)=f(0)+e^{0}\sqrt{0+1}=2+1=3$

$(f^{-1})'(2)=\dfrac{1}{f\,'(0)}=\dfrac{1}{3}$

$\Rightarrow \text{option } (2)$