Question

If $f :[1, \infty) \rightarrow[\sqrt{2}, \infty]$ be a function defined by $f(x)=\sqrt{x^2-2 x+3}$ and $g$ be the inverse function of $f$ then derivative of $g\left(f^2(x)\right)$ at $x=3$ is

• A. $\frac{6}{\sqrt{34}}$
• B. $\frac{\sqrt{34}}{6}$
• C. $\frac{24}{\sqrt{34}}$
• D. $\frac{12}{\sqrt{34}}$

Answer 1

Answer: (C)

$g\left( f ^2( x )\right)= g \left( x ^2-2 x +3\right)$
Now, $\frac{d}{d x}\left(g\left(x^2-2 x+3\right)\right)=g^{\prime}\left(x^2-2 x+3\right)(2 x-2)$
$\left.\therefore \frac{ d }{ dx }\left( g \left( f ^2( x )\right)\right)\right|_{ x =3}= g ^{\prime}(6) \cdot 4=4 g ^{\prime}(6)$
Now, $ g^{\prime}(6)=\frac{1}{f^{\prime}(1+\sqrt{34})}$
$f ^{\prime}( x )=\frac{2( x -1)}{2 \sqrt{( x -1)^2+2}} \Rightarrow f ^{\prime}(1+\sqrt{34})=\frac{\sqrt{34}}{6} $
$\therefore g ^{\prime}(6)=\frac{6}{\sqrt{34}} \Rightarrow 4 g ^{\prime}(6)=\frac{24}{\sqrt{34}}$