If $g$ is the inverse of $f$ and $f'(x) = \frac{1}{1 + x^5}$,then $g'(x) =$

Let $f: \{1, 2, 3\} \rightarrow \{a, b, c\}$ be a one-one and onto function given by $f(1) = a$,$f(2) = b$,and $f(3) = c$. Show that there exists a function $g: \{a, b, c\} \rightarrow \{1, 2, 3\}$ such that $g \circ f = I_X$ and $f \circ g = I_Y$,where $X = \{1, 2, 3\}$ and $Y = \{a, b, c\}$.

If the function $f : R \to R$ is defined by $f(x) = \log_a(x + \sqrt{x^2 + 1})$,where $(a > 0, a \neq 1)$,then $f^{-1}(x)$ is

Let $f: R - \{\frac{\alpha}{6}\} \rightarrow R$ be defined by $f(x) = \frac{5x + 3}{6x - \alpha}$. Then the value of $\alpha$ for which $(f \circ f)(x) = x$,for all $x \in R - \{\frac{\alpha}{6}\}$,is:

If $f(x) = \frac{3x+2}{5x-3}$,where $x \in R - \{\frac{3}{5}\}$,then:

Let $g(x)$ be the inverse of the function $f(x)$ and $f'(x) = \frac{1}{1 + x^3}$. Then $g'(x)$ is equal to