Let g(x) be the inverse of the function f(x) and f'(x) = $\frac{1}{1+x^3}$. Then g'(x) is :

$\frac{1}{1+(g(x))^3}$ $\frac{1}{1+(f(x))^3}$ $1+(g(x))^3$ $1+(f(x))^3$

$1+(g(x))^3$

Since g(x) is the inverse of f(x) ∴ (gof)(x) = x ⇒ g(f(x)) = x ⇒ g'(f(x)) f'(x) = 1 ⇒ $g'(f(x))=\frac{1}{f'(x)}=1+x^3$ ⇒ g'(y) = 1 + (g(y))2 [∵ y = f(x) ⇒ x = f–1(y) = g(y)] ⇒ g'(x) = 1 + (g(x))3.        (Replacing y by x) Hence (3) is correct answer.