Question

If $f\left(x\right)=\ln (x^2+2x+2)$ and $g$ is the inverse of function $f,$ then the value of $g^{\prime }\left(\ln 2\right)$ is

Answer 1

Given :
$f\left(x\right)=\ln (x^2+2x+2)\Rightarrow f\left(0\right)=\ln 2$
Differentiating both sides w.r.t. $x$
$f^{\prime }\left(x\right)=\frac{1}{x^2+2x+2}\times (2x+2)$
$\Rightarrow f^{\prime }\left(0\right)=1$
and $f^{-1}\left(x\right)=g\left(x\right)\Rightarrow g^{-1}\left(x\right)=f\left(x\right)$
$\therefore g\left(f\right(x\left)\right)=x$
Differentiating both sides w.r.t. $x$
$g^{\prime }\left(f\right(x\left)\right)\cdot f^{\prime }\left(x\right)=1$
$\Rightarrow g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}$
$\Rightarrow g^{\prime }\left(f\right(0\left)\right)=\frac{1}{f^{\prime }\left(0\right)}$
$\therefore g^{\prime }\left(\ln 2\right)=\frac{1}{f^{\prime }\left(0\right)}=1$