Question

If $g\left(x\right)$ is the inverse of $f\left(x\right)$ and $f^{\prime }\left(x\right)=\frac{1}{1+x^3}$, then $g^{\prime }\left(x\right)$ is equal to

• A. $g\left(x\right)$
• B. $1+g\left(x\right)$
• C. $1+{\left\{g\right(x\left)\right\}}^3$
• D. $\frac{1}{1+{\left\{g\right(x\left)\right\}}^3}$
• E. $0$

Answer 1

Answer: (C)

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

Given, $g\left(x\right)=f^{-1}\left(x\right)$

$\Rightarrow f\left\{g\right(x\left)\right\}=x$

On differentiating, w.r.t. $x$, we get

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

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

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

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

Now, from Eq. (i), we get

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