so before i start this is extracted from spivak's calculus proof(adsbygoogle = window.adsbygoogle || []).push({});

now here is the problem the teacher who taught us limits and dervatives for the first time told us not to use f(x)-f(a) instead of f(a+h)-f(a)

so spivak wanted us to prove that if f a one-one function and is continious on an interval and differentiable on f[itex]^{-1}[/itex](b) then f'(b) exists and (f^-1)'(b)=[itex]\frac{1}{f'(f^-1(b))}[/itex](f^-1 = inverse of f )

i did the opposite of what spivak did , spivak started from the definition of f'(b) while i started wtih the definition of f[itex]^{-1}[/itex](b), so is my proof right?

let f(a) = b then f^-1(b)=a so

f'(f[itex]^{-1}[/itex](b) = f'(a) = [itex]\lim_{x \to a}\frac{f(x)-f(a)}{x-a} [/itex]=[itex]\lim_{x \to f^-1(b)}\frac{f(x)-b}{x-f^-1(b)} [/itex]

now when x approches a=f^-1(b) in the domain of f , f(x)approaches f(a)=b in the domain on f-1 and we can set f(x)=x later on

so =[itex]\lim_{f(x) \to b}\frac{f(x)-b}{x-f^-1(b)} [/itex]

replacing f(x) by the new x in the domain of f^-1 and b by f^-1(b) and x by f^-1(x)= a(I hope i'm right in this phase, please correct me if i'm wrong)

=[itex]\lim_{x \to b}\frac{x-b}{f^-1(x)-f^-1(b)} [/itex] = [itex]\frac{1}{(f^-1)'(b)}[/itex]

i've always been confused when changing the x to f(x) : do we also change the function or it remains intact?Limx->af(x) <=>Limf(x)->f(a) [f(f(x))) or f(x)? ] if i made a fault it would be this, thanks for reading.

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# Can't decide if my proof is right or not ?inverse functions

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