Question

If g is inverse of f and f'(x)=1/1+x^3
then find g'(x).

Answer 1

@zepdrix @experimentX

Answer 2

whadoyou mean??
\[ f^{-1}(x) = \frac{1}{1+x^3}\]
and \[ g(x) = f^{-1}(x)\]
??

Answer 3

yes correct

Answer 4

then
\[ g(x) = \frac{1}{1+x^3}\]
find the derivative ... what's the problem?

Answer 5

um sorry its f'(x) not f inverse x

Answer 6

straight way, this looks bad
http://www.wolframalpha.com/input/?i=Integrate+1%2F%281%2Bx%5E3%29

Answer 7

we are not supposed to integrate anything here..

Answer 8

Derivative of the Inverse Function:
\[\large \frac{d}{dx}f^{-1}(x)=\frac{1}{f^{-1}\left[f'(x)\right]}\]


I think we can use this somehow..

\[\large g^{-1}(x)=f(x)\]

\[\large \frac{d}{dx}g^{-1}(x)\qquad=\qquad\frac{d}{dx}f(x)\qquad=\qquad\frac{1}{f^{-1}\left[f'(x)\right]}\]

Hmm I'm not doing this correctly... but I think this problem will involve that formula somehow... thinkinggg

Answer 9

Oh i wrote that incorrectly,\[\large \frac{d}{dx}g(x)=\frac{d}{dx}f^{-1}(x)=...\]Yah i think this will work.

Answer 10

\[\large =\frac{1}{f^{-1}\left[f'(x)\right]}\]

Answer 11

How about this??
Multiplying f(x) on both sides..
\[\large f(g(x))=x\]
Differentiating..
\[\Large f'(g(x)).g'(x)=1\]
\[\Large g'(x)=1+g(x)^3\]

Answer 12

is 0 on option?

Answer 13

\[\large g'(x)=\frac{1}{f^{-1}\left[\dfrac{1}{1+x^3}\right]}\]

Answer 14

nope,ans is w hat i wrote

Answer 15

Hmm

Answer 16

yes yes ... sorry, i equated with 1

Answer 17

thanks! :)