Question

Let $f$ be a function defined implicitly by the equation $\frac{1-e^{f\left(x\right)}}{1+e^{f\left(x\right)}}=x$ and $g$ be the inverse of $f.$ If $g^{\prime \prime }\left(\ln 3\right)-g^{\prime }\left(\ln 3\right)=\frac{p}{q},$ where $p$ and $q$ are relatively prime, then the value of $(p+q)$ is

Answer 1

Given, $\frac{1-e^{f\left(x\right)}}{1+e^{f\left(x\right)}}=x$
$f$ and $g$ are inverse of each other.
So, $g\left(f\right(x\left)\right)=x$
$\Rightarrow \frac{1-e^{f\left(x\right)}}{1+e^{f\left(x\right)}}=g\left(f\right(x\left)\right)$
$\therefore g\left(x\right)=\frac{1-e^x}{1+e^x}$

$g^{\prime }\left(x\right)=\frac{-2e^x}{(1+e^x{)}^2}=-2e^x(1+e^x{)}^{-2}$
$g^{\prime \prime }\left(x\right)=-2\left[e^x\right(1+e^x{)}^{-2}+e^x(-2)(1+e^x{)}^{-3}\cdot e^x]$
$=-2\left[e^x\right(1+e^x{)}^{-2}-2(1+e^x{)}^{-3}\cdot (e^x{)}^2]$

Now, $g^{\prime }\left(\ln 3\right)=\frac{-3}{8}$
and $g^{\prime \prime }\left(\ln 3\right)=\frac{3}{16}$
$\therefore g^{\prime \prime }\left(\ln 3\right)-g^{\prime }\left(\ln 3\right)=\frac{3}{16}+\frac{3}{8}=\frac{9}{16}$
Hence, $p+q=9+16=25$