Question

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

• A. $\cot x$
• B. $\frac{1}{1+x^2}$
• C. $\frac{1}{1-x^2}$
• D. $\tan x$

Answer 1

Answer: (B)

$\frac{1}{1+x^2}$

Given $f\left(x\right)=\tan x$ and $f$ is inverse of $g$

So , $g\left(x\right)$ is ${\tan }^{-1}x$

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