Explanation

(c)
Since, $\phi(x) = f^{- 1}(x)\ \ \ \Rightarrow \ \ \ \ \ x = f\{\phi(x)\}$
On differentiating w.r.t. $x,\$we get
$1 = f^{'}\left\{ \phi(x) \right\}.\phi^{'}(x)$
$\Rightarrow$ $\phi^{'}(x) = \frac{1}{f^{'}\{\phi(x)\}}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(i)$
But $f^{'}\left\{ \phi(x) \right\} = \frac{1}{1 + \left\{ \phi(x) \right\}^{5}}\ \ \ \ \ \left( \because f^{'}(x) = \frac{1}{1 + x^{5}} \right)$
$\therefore From\ Eq.(i),$
$\phi^{'}(x) = \frac{1}{f^{'}\{\phi(x)\}} = 1 + \left\{ \phi(x) \right\}^{5}$