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Q.
If $g$ is inverse of $f$ and $f( x )= x ^2+3 x -3( x >0)$ then $g ^{\prime}(1)$ equals-
Continuity and Differentiability
Solution:
$f ^{-1}( x )= g ( x ) \Rightarrow x = f ( g ( x ))$
Differentiating both sides,
$1=f^{\prime}(g(x)) g^{\prime}(x) \Rightarrow g^{\prime}(x)=\frac{1}{f^{\prime}(g(x))}$
Now $f ^{\prime}( x )=2 x +3$
So $g^{\prime}(x)=\frac{1}{2 g(x)+3} \Rightarrow g^{\prime}(1)=\frac{1}{2(g(1))+3}$
$\text{gof}( x )= x \Rightarrow g ^{\prime}( f ( x )) f ^{\prime}( x )=1$
$f(x)=1 $ at $x=1 \& f^{\prime}(1)=5$
$g^{\prime}(1) f^{\prime}(1)=1 \Rightarrow g^{\prime}(1)=1 / 5$