Question

If $f(x)=x^3+3x+4$ and $g$ is the inverse function of $f$ then the value of $\frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right)$ at $x=4$ equals

• A. $\frac{-1}{6}$
• B. 6
• C. $\frac{-1}{3}$
• D. non-existent

Answer 1

Answer: (C)

$\frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right)$ at $x=4$
${=\frac{g(g(x))g^{\prime }(x)-g(x)\cdot g^{\prime }(g(x))\cdot g^{\prime }(x)}{(g(g(x)){)}^2}]}_{x=4}$
Now, $f(0)=4\Rightarrow f^{-1}(4)=g(4)=0$ and $g^{\prime }(4)=\frac{1}{f^{\prime }(0)}=\frac{1}{3}$
$f(-1)=0\Rightarrow f^{-1}(0)=g(0)=-1$
$=\frac{g(g(4))g^{\prime }(4)-g(4)g^{\prime }(g(4))g^{\prime }(4)}{(g(g(4)){)}^2}=\frac{g(0)\cdot \frac{1}{f^{\prime }(0)}-0}{(g(0){)}^2}=\frac{-1\times \frac{1}{3}}{1}=\frac{-1}{3}$