If `f(x)=x+tanxa n df` is the inverse of `g,` then `g^(prime)(x)` equals `1/(1+[g(x)-x]^2)` (b) `1/(2-[g(x)-x]^2)` `1/(2+[g(x)-x]^2)` (d) none of these

Options

A. `(1)/(1+[g(x)-x]^(2))`
B. `(1)/(2-[g(x)-x]^(2))`
C. `(1)/(2+[g(x)-x]^(2))`
D. none of these

Answer

C

Explanation

`f(x)=x+tan x`
`therefore" "f(f^(-1)(y))=f^(-1)(y)+tan f^(-1)(y)`
`"or "y=g(y)+tan g(y)`
`"or "x=g(x)+tan g(x)`
Differentiating, we get `1=g'(x)+sec^(2)g(x)g'(X)`
`"or "g'(x)=(1)/(1+sec^(2)g(x))=(1)/(2+[g(x)-x]^(2))`