Since $f(x)$ and $g(x)$ are inverse of each other $\therefore$ $g'\left( {f\left( x \right)} \right) = {1 \over {f'\left( x \right)}}$ $\Rightarrow g'\left( {f\left( x \right)} \right) = 1 + {x^5}$ $\left( \, \right.$ As $\,f'\left( x \right) = {1 \over {1 + {x^5}}}$ $\left. \, \right)$ Here $x=g(y)$ $\therefore$ $g'\left( y \right) = 1 + \left\{ {g\left( y \right)} \right\}$ $\Rightarrow g'\left( x \right) = 1 + \left\{ {g\left( x \right)} \right\}$