1. ## Derivtive of Inverse

Hey guys Im not sure how to do this and would be delighted with any help. I was thinking simplifying (f -1 o f)(x) and then chain rule.

Suppose that (
a, b) to(c, d) is a bijection,
and that f is differentiable on (a, b). Assuming that this implies that the inverse function
f -1 : (c, d) to (a, b) is differentiable on (c, d), show that if y0 element of(c; d) and x0= f -1(y0) then,

(f -1)'(y0) = 1/f '((f -1)(y0)), if f '(x0) is not 0

2. ## Re: Derivtive of Inverse

Originally Posted by MathJack
Hey guys Im not sure how to do this and would be delighted with any help. I was thinking simplifying (f -1 o f)(x) and then chain rule.

Suppose that (a, b) to (c, d) is a bijection,and that f is differentiable on (a, b). Assuming that this implies that the inverse function
f -1 : (c, d) to (a, b) is differentiable on (c, d), show that if y0 element of (c; d) and x0 = f -1(y0) then,

(f -1)'(y0) = 1/f '((f -1)(y0)), if f '(x0) is not 0

Recall that $\displaystyle f^{-1}(f(x))=x$ thus $\displaystyle {D_x}\left[ {{f^{ - 1}}\left( {f(x)} \right)} \right] = \left[ {{{\left( {{f^{ - 1}}} \right)}^\prime }\left( {f(x)} \right)} \right] \cdot f'(x) = 1$.

So if $\displaystyle f(x_0)=y_0$ we get $\displaystyle {\left( {{f^{ - 1}}} \right)^\prime }\left( {f({x_0})} \right) = {\left( {{f^{ - 1}}} \right)^\prime }\left( {{y_0}} \right) = \frac{1}{{f'({x_0})}}$