If $g$ is the inverse of $f$ and $f^{\prime}(x)=\frac{1}{1+x^{2}}$,then $g^{\prime}(x)$ is equal to

Let $f: R-\{2\} \rightarrow R-\{1\}$ defined by $f(x)=\frac{x-3}{x-2}$ and $g: R \rightarrow R$ defined by $g(x)=3x-2$. Then,the sum of all values of $x$ for which $f^{-1}(x)+g^{-1}(x)=\frac{19}{6}$ is

If a function $f:(-1,1) \rightarrow B(\subseteq R)$ is defined as $f(x)=x+x^2+x^3+\ldots \infty$, then in order to have the inverse function of $f$, $B$ is equal to

If $y = f(x) = \frac{ax + b}{cx - a}$,then $x$ is equal to

If $f(x) = (2x - 3\pi)^5 + \frac{4}{3}x + \cos x$ and $g$ is the inverse of $f$,then $g'(2\pi) = ?$

If $f(x) = x^2 - x + 5, x > \frac{1}{2},$ and $g(x)$ is its inverse function,then $g'(7)$ equals