`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))`

Options

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)))`

Answer

C

Explanation

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))"`