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Q.
If $f\left(x\right)=x^{3}+3x+4$ and $g$ is the inverse function of $f,$ then the value of $\frac{d}{d x}\left(\frac{g(x)}{g(g(x))}\right)$ at $x=4$ equals:-
NTA AbhyasNTA Abhyas 2022
Solution:
Let $D=\frac{d}{d x}\left(\frac{g(x)}{g(g(x))}\right)$ at $x =4$
$
\left.=\frac{ g ( g ( x )) g ^{\prime}( x )- g ( x ) \cdot g ^{\prime}( g ( x )) \cdot g ^{\prime}( x )}{( g ( g ( x )))^{2}}\right]_{ x =4}
$
Now, $f(x)=x^{3}+3 x+4$
$
\Rightarrow f(x)=3 x^{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(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)=-1
$
$
\therefore D =\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}
$