## Calculus 8th Edition

$(f^{-1})'(0)=\frac{1}{\sqrt 28}$
Let $x=3$, then $f(3)=\int\sqrt {1+t^3}dt$=0, so $f^{-1}(0)=3$. Furthermore, $f'(x)=\sqrt{1+t^3}$. Now, by theorem 7, $(f^{-1})'(0)=\frac{1}{f'(f^{-1}(0))}=\frac{1}{f'(3)}=\frac{1}{\sqrt {28}}.$