Suppose that f has an inverse, f(3)=1 and f'(3)=5/3. What is (f^-1)'(1)?
$\displaystyle f(f^{-1}(x))=x$
Take derivative of both sides,
$\displaystyle [f^{-1}(x)]'f'(f^{-1}(x))=1$
At $\displaystyle x=1$ we have,
$\displaystyle [f^{-1}(1)]'f'(f^{-1}(1))=1$
$\displaystyle [f^{-1}(1)]'f'(3)=1$
$\displaystyle [f^{-1}(1)]'\cdot \frac{5}{3}=1$.