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`f` is a strictly monotonic differentiable function with `f^(prime)(x)=1/(sqrt(1+x^3))dot` If `g` is the inverse of `f,` then `g^(x)=` `(2x^2)/(2sqrt(1+x^3))` b. `(2g^2(x))/(2sqrt(1+g^2(x)))` c. `3/2g^2(x)` d. `(x^2)/(sqrt(1+x^3))`

A

`(3x^(2))/(2sqrt(1+x^(3)))`

B

`(3g^(2)(x))/(2sqrt(1+g^(2)(x)))`

C

`(3)/(2)g^(2)(x)`

D

`(x^(2))/(sqrt(1+x^(3)))`

Text Solution

Verified by Experts

The correct Answer is:
C

g is inverse of f
`therefore" "fog(x)=x`
`therefore" "f'(g(x)).g'(x)=1`
`therefore" "g'(x)=(1)/(f'(g(x)))" (1)"`
`=sqrt(1+g^(3)(x))`
`therefore" "g''(x)=(3g^(2)(x).g'(x))/(2sqrt(1+g^(3)(x)))`
`=(3g^(2)(x))/(2)" (using (1))"`
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