Question

How do I solve this?
Find (f^-1)'(a) if f(x)=(x^3)+1 and a=9

Answer 1

\[D_xf^{-1}(x)=\frac{1}{f'(f^{-1}(x))}\]

Answer 2

what is\[f^{-1}(x)\]what is\[f'(x)\]what is\[(f'of^{-1})(x)\]...

Answer 3

The inverse is \[\sqrt[3]{y+1}\] and the derivative is \[3x^{3}\] so the answer should be \[1/3(\sqrt[3]{y+1})^{3}\] which gives me 1/3(9+1) which is 1/30... but the answer is 1/12 what did I do wrong?

Answer 4

you made mistakes, but still there is no 12 in the problem :/

Answer 5

inverse is\[f^{-1}(x)=(x-1)^{1/3}\]derivative is\[f'(x)=3x^2\]so we have\[\frac{d}{dx}f^{-1}(x)=\frac{1}{3(x-1)^{2/3}}\]do remember to change the y's back to x's for the inverse of f(x).
as I said though, no 1/12...