- #1
Bacle
- 662
- 1
Problem with proof of Chain rule for f:R-->R
Hi, Analysts:
I am going over problems in Rosenlicht's Intro. Analysis
book. In this problem , he asks one to find the flaw in this
argument to the effect that (f(g(x))'=f'(g(x))g'(x). Unfortunately,
author does not clearly state the assumptions; I believe author assumes
f differentiable at g(xo), and g'(xo) exists.
Here is the exercises:
(f(g(xo))=
1) Lim_x->xo [f(g(xo))-f(g(x)]/(x-xo)=
(multiply by (g(x)-g(xo))/(g(x)-g(xo))
2) Lim_x->xo [[f(g(xo)-f(g(x)]/(g(x)-g(xo) ]*[(g(x)-g(xo)]/(x-xo)=
3) Lim_x->xo [f(g(xo)-f(g(x))/(g(x)-g(xo)]*Lim_x->xo[(g(x)-g(xo)/(x-xo)]=
4) f'(g(xo))*g'(xo).
I know the problem is in step 3, since f(g(xo)-f(g(x)) can be zero, even for f
continuous ( or, even smooth, real-analytic, etc.) , using examples like
f(x)=x^2 sin(1/x) for x=/0 and f(x)=0 otherwise.
My question is: what additional conditions do we need on g to make the proof
work, and/or how to change the proof to make it work for a more general case.?
Thanks.
Hi, Analysts:
I am going over problems in Rosenlicht's Intro. Analysis
book. In this problem , he asks one to find the flaw in this
argument to the effect that (f(g(x))'=f'(g(x))g'(x). Unfortunately,
author does not clearly state the assumptions; I believe author assumes
f differentiable at g(xo), and g'(xo) exists.
Here is the exercises:
(f(g(xo))=
1) Lim_x->xo [f(g(xo))-f(g(x)]/(x-xo)=
(multiply by (g(x)-g(xo))/(g(x)-g(xo))
2) Lim_x->xo [[f(g(xo)-f(g(x)]/(g(x)-g(xo) ]*[(g(x)-g(xo)]/(x-xo)=
3) Lim_x->xo [f(g(xo)-f(g(x))/(g(x)-g(xo)]*Lim_x->xo[(g(x)-g(xo)/(x-xo)]=
4) f'(g(xo))*g'(xo).
I know the problem is in step 3, since f(g(xo)-f(g(x)) can be zero, even for f
continuous ( or, even smooth, real-analytic, etc.) , using examples like
f(x)=x^2 sin(1/x) for x=/0 and f(x)=0 otherwise.
My question is: what additional conditions do we need on g to make the proof
work, and/or how to change the proof to make it work for a more general case.?
Thanks.