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My version of theorem (to prove):

Let $f:(a,b)\to (c,d)$ be an invertible continuous function, (read carefully) differentiable at $m\in(a,b)$, such that $f'(m)\neq0$. Then $(f^{-1})'(f(m))=\frac1{f'(m)}$.

For convenience, let $g(x)=f^{-1}(x)$.

We have to prove that for every $\epsilon>0$, there exists a $\delta>0$, such that, $$0<|y-f(m)|<\delta \implies|\frac{g(y)-m}{y-f(m)}-\frac{1}{f'(m)}|<\epsilon$$ Now, note that in the above expression, $y\in(c,d)$. Also, for every $y\in(c,d)$ we have a unique $x\in(a,b)$ (and vice versa). ($y=f(x)$)

So above statement can be written as, we have to prove that for every $\epsilon>0$, there exists $\delta>0$, such that, $$0<|f(x)-f(m)|<\delta\implies|\frac{x-m}{f(x)-f(m)}-\frac{1}{f'(m)}|<\epsilon$$

Now since, $f(x)$ is continuous (and injective (think what is the need of one-oneness)), for any positive number $\delta$, we have another number $\delta'$, so that $$0<|x-m|<\delta'\implies 0<|f(x)-f(m)|<\delta$$

So we have to prove that $\forall$ $\epsilon>0$ there exists a $\delta'>0$, so that. $$0<|x-m|<\delta'\implies|\frac{x-m}{f(x)-f(m)}-\frac{1}{f'(m)}|<\epsilon$$ i.e. $$\lim_{x\to m} \frac{x-m}{f(x)-f(m)}=\frac{1}{f'(m)}$$ Which is true by limit theorem.

So is my version of theorem, and the proof, correct?

Aditya Agarwal
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  • Two things that bother me:1) If you write $(f^{-1}(f(x))'$ you first evaluate the composition of the function and derive afterwards. Hence $(f^{-1}(f(x))'=(x)'=1$. I think it would be better to write $(f^{-1})'(f(x))$ instead. The other one would be that the logic of your argument is inverted. You start at what you want to show and deduce the assumption. It has to be the other way around, i.e. you have to start at $\lim_{x\rightarrow m}\frac{x-m}{f(x)-f(m)}=\frac{1}{f'(m)}$. – humanStampedist Sep 18 '18 at 08:58
  • @humanStampedist For the first point, thanks! But as far as the second one is concerned, I think my deductions are valid. I just started by showing that what I want to prove is equivalent to prove the next, and then finally I reach a point, which is equivalent to the first to-prove statement, and I prove it. – Aditya Agarwal Sep 18 '18 at 09:13
  • I do not see, that your deductions are equivalent statements. For that to work you need, that $f^{-1}$ is continuous (see https://math.stackexchange.com/questions/1523622/continuous-function-with-compact-domain-has-continuous-inverse) so that you can see that if $|f(x)-f(m)|$ is small that then $|x-m|$ is small too. – humanStampedist Sep 18 '18 at 09:16
  • @humanStampedist Nice! Gotcha! – Aditya Agarwal Sep 18 '18 at 09:21

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