Question

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

• A. $\frac{1}{1+\left(g\right(x){)}^3}$
• B. $\frac{1}{1+\left(f\right(x){)}^3}$
• C. $1+\left(g\right(x){)}^3$
• D. $1+\left(f\right(x){)}^3$

Answer 1

The correct option is C $1+\left(g\right(x){)}^3$

Given that $g\left(x\right)$ is inverse of $f\left(x\right)$.

$\Rightarrow g\left(f\right(x\left)\right)=x\Rightarrow g^{\prime }\left(f\right(x\left)\right).f^{\prime }\left(x\right)=1$

$g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}=1+x^3$

$f\left(x\right)=y$

$g^{\prime }\left(y\right)=\frac{1}{f^{\prime }\left(x\right)}=1+x^3$

$x=f^{-}1\left(y\right)=g\left(y\right)$..............................................................(i)

$g^{\prime }\left(y\right)=1+g^3\left(y\right)$ by (i)

$g^{\prime }\left(x\right)=1+g^3\left(x\right)$