Question

If $f:[1,\infty )\to [\sqrt{2},\infty ]$ be a function defined by $f(x)=\sqrt{x^2-2x+3}$ 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)

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