Question

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

• A. $\frac{1}{1+\left[g\right(x)-x{]}^2}$
• B. $\frac{1}{2-\left[g\right(x)+x{]}^2}$
• C. $\frac{1}{2+(x-g(x){)}^2}$
• D. $Noneofthese$

Answer 1

Answer: (C)

$\frac{1}{2+(x-g(x){)}^2}$

Since,

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

$\Rightarrow f\left(g\right(x\left)\right)=g\left(x\right)+\tan \left(g\right(x\left)\right)=x$

$\tan \left(g\right(x\left)\right)=x-g\left(x\right)$

Differentiating both sides with respect to $x$

$g^{\prime }\left(x\right)+{\sec }^2\left(g\right(x\left)\right)g^{\prime }\left(x\right)=1$

$g^{\prime }\left(x\right)=\frac{1}{1+{\sec }^2\left(g\right(x\left)\right)}=\frac{1}{2+{\tan }^2\left(g\right(x\left)\right)}$

$g^{\prime }\left(x\right)=\frac{1}{2+(x-g(x){)}^2}$

Hence, option $C$ is correct answer.