Implicit Differentiation Calculus 1 Question

Question

assume that y=f(x) has an inverse on an interval around x = a and that f'(a) = 0 why does f^-1(y) not have a derivative at y = f(a)?

Answer 1

The derivative of f^-1 at f(a) is the reciprocal of the derivative of f(x) at x = a.

So, since f'(a) = 0, the derivative of f^-1 at f(a) is 1/0, which is undefined.

Answer 2

Let g(x) = f^-1(x)

By definition g(f(x)) = x

and taking the derivative of both sides of the equation and using the chain rule

g'(f(x))f'(x) = 1

So g'(f(x)) = 1/f'(x)

If f'(a) = 0 then g'(f(a)) = 1/f'(a) which is undefined since f'(a) = 0.