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 $x=4$ is

• A. $\frac{-1}{3}$
• B. $\frac{-1}{2}$
• C. $3$
• D. $6$

Answer 1

Answer: (A)

Let $D=\frac{d}{dx}\left(\frac{g(x)}{g(g(x))}\right)$ at $x=4$
${=\frac{g(g(x))g^{\prime }(x)-g(x)\cdot g^{\prime }(g(x))\cdot g^{\prime }(x)}{(g(g(x))}]}_{x=4}$
Now, $f(x)=x^3+3x+4$
$\Rightarrow f^{\prime }(x)=3x^2+3>0$
Clearly, $f(x)$ is an increasing function.
Now, $f(0)=4\Rightarrow f^{-1}(4)=g(4)=0$
Also, $g(f(x))=x$
$\begin{array}{l}\therefore g^{\prime }(f(x))f^{\prime }(x)=1\\ \Rightarrow g^{\prime }(f(0))f^{\prime }(0)=1\\ \Rightarrow g^{\prime }(4)=\frac{1}{f^{\prime }(0)}=\frac{1}{3}\end{array}$
$f(-1)=0\Rightarrow f^{-1}(0)=g(0)=-11$
$\therefore D=\frac{g(g(4))g^{\prime }(4)-g(4)g^{\prime }(g(4))g^{\prime }(4)}{(g(g(4))}$
$=\frac{g(0)\cdot \frac{1}{f^{\prime }(0)}-0}{(g(0)}=\frac{(-1)\times \frac{1}{3}}{1}=\frac{-1}{3}$