Consider $f(x) = x^3 - 3x^2 - 1$ where $x \geq 2$.
Find derivative for $f^{-1}(x)$ at point where $x = -1 = f(3)$.
I tried getting $f^{-1}(x)$ and then getting the derivative of that but it th point equal. I also tried getting $f'(x)$ and inverting that was wrong too.
$$ f (x)=x^3-3x^2-1 \ \ \ \implies \ \ \ f '(x)= 3x^2-6x \ \ \ \ \ \ \implies \ \ \ \ f '(3) = 9 $$ let : $$ f^{−1} = g $$ so that$$ g(f\:(x))=x\ \ \ \ (\because\ \ g= f^{-1})$$
differentiating wrt x and chain rule:
$ g'(f(x)).f'(x)=1 $
$ \implies g'(f(x))\ =\frac{1}{f'(x)} $
putting $x=3 \ \implies \ \ \ f(3)=-1 $
$\therefore$ derivative of $g$ at $ x=-1 $
$g'(-1)=\frac{1}{f'(3)}=\frac{1}{9} $
now,
$$g = f^{-1} $$ thus,
$$ \frac{d}{dx} \left(f^{-1}(-1)\right)=\frac{1}{9}$$
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