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

Question

turingtest · Answer

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

turingtest · Answer

what is$$f^{-1}(x)$$what is$$f'(x)$$what is$$(f'of^{-1})(x)$$...

anonymous · Answer

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?

turingtest · Answer

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

turingtest · Answer

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...