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Q.
If $f(x)=x^3+3 x+4$ 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
Continuity and Differentiability
Solution:
$\frac{ d }{ dx }\left(\frac{ g ( x )}{ g ( g ( x ))}\right)$ 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, $f (0)=4 \Rightarrow f ^{-1}(4)= g (4)=0$ and $g ^{\prime}(4)=\frac{1}{ f ^{\prime}(0)}=\frac{1}{3}$
$f(-1)=0 \Rightarrow f^{-1}(0)=g(0)=-1$
$=\frac{g(g(4)) g^{\prime}(4)-g(4) g^{\prime}(g(4)) g^{\prime}(4)}{(g(g(4)))^2}=\frac{g(0) \cdot \frac{1}{f^{\prime}(0)}-0}{(g(0))^2}=\frac{-1 \times \frac{1}{3}}{1}=\frac{-1}{3}$