Let f(x) = x ^ 2 - 4x - 3,x > 2 and g ( x ) be the inverse of f ( x...

As& $g ( f ( x ) ) = x$~$( ∵ g ( x ) = f^{- 1} ( x ) )$
Differentiating, we get,~

$g^{'} ( f ( x ) ) f^{'} ( x ) = 1$

$⇒g^{'} ( f ( x ) ) = \dfrac{1}{f^{'} ( x )}$

when~$f ( x ) = 2$

$⇒x^{2} - 4x - 3 = 2$

$⇒x = 5, x = - 1$

$∵ x > 2$

$∴ x = 5$

$g^{'} ( 2 ) = \dfrac{1}{( 2x - 4 )x = 5}$

$= \dfrac{1}{6}$