1
$\begingroup$

I have tried using the definition of derivative by

$$ \lim_{h \to 0} \dfrac{f^{-1}\left(x + h\right) - f^{-1}\left(x\right)}{h} $$

but that is not correct. (it was marked wrong).

What did I do wrong?

  • 1
    Why do you think that is not correct? I'd say it is, assuming the limit exists.2012-11-07
  • 0
    Probably, they wanted to express $(f^{-1})'$ using $f'$ and $f$..2012-11-07
  • 0
    @berci I think you are correct.2012-11-08

1 Answers 1

3

This is correct so far, but you should go on, somehow introducing the definition of $f'$.

Briefly, it goes like $t:=f^{-1}(x+h)-f^{-1}(x)$, we need that $t\to 0$ as $h\to 0$, and then consider $y:=f^{-1}(x)$ and $$ h = (x+h)-x = f(y+t) -f(y) $$

  • 0
    thanks, I will try to work that path out.2012-11-08