Suppose f^-1 is the inverse function of a differentiable function f and f(4)=9 , f'(4)=7/5, then (f^-1)'(9)=?

Answer 1

Here is a short sketch of solving that

{[d(f^-1)/dy] at a point y=y_o ]}=[1/(df/dx) where f(x)=y_o]

Here y_o=9 ; f(4)=9; df/dx=7/5 at x=4

Therefore d(f^-1)/dy at y=9 would be equal to

1/(7/5)=5/7