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. 3

Answer 1

Answer: (C)

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