Let `phi (x)` be the inverse of the function `f(x) and f'=1/(1+x^5),` then `d/(dx) phi(x)` is

Options

A. `(1)/(1+[phi(x)]^(5))`
B. `(1)/(1+[f(x)]^(5))`
C. `1+[phi(x)]^(5)`
D. `1+f(x)`

Answer

C

Explanation

Given `phi(x)=f^(-1)(x)`
`impliesx=f[phi(x)]`
On differentiating both sides w.r.t. x, we get
`1=f'[phi(x)]*phi'(x)`
`impliesphi'(x)=(1)/(f'[phi(x)])` . . . (i)
but `f'(x)=(1)/(1+x^(5))`
`thereforef'[phi(x)]=(1)/(1+[phi(x)]^(5))`
From Eq. (i) `phi'(x)=1+[phi(x)]^(5)`