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

Let $R$ denote the set of all real numbers. Let $f: R \rightarrow R$ and $g: R \rightarrow (0, 4)$ be functions defined by $f(x) = \log_e(x^2 + 2x + 4)$ and $g(x) = \frac{4}{1 + e^{-2x}}$. Define the composite function $h(x) = (f \circ g^{-1})(x)$,where $g^{-1}$ is the inverse of the function $g$. Then the value of the derivative of the composite function $h(x)$ at $x = 2$ is:

Consider $f: R_{+} \rightarrow [4, \infty)$ given by $f(x) = x^{2} + 4$. Show that $f$ is invertible with the inverse $f^{-1}$ of $f$ given by $f^{-1}(y) = \sqrt{y - 4}$,where $R_{+}$ is the set of all non-negative real numbers.

Consider $f: R_{+} \rightarrow [-5, \infty)$ given by $f(x) = 9x^{2} + 6x - 5$. Show that $f$ is invertible with $f^{-1}(y) = \frac{\sqrt{y+6}-1}{3}$.

If $f(x) = \frac{x^2-x}{x^2+2x}$,then the value of $\frac{d}{dx}(f^{-1}(x))$ at $x = 2$ is:

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