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Q.
Let $g$ is the inverse function of $f_{\text {and }} f^{\prime}(x)=\frac{x^{10}}{\left(1+x^2\right)}$. If $g(2)=a$ then $g^{\prime}(2)$ is equal to
Continuity and Differentiability
Solution:
$f [ g ( x )]= x \Rightarrow f ^{\prime}[ g ( x )] \cdot\left[ g ^{\prime}( x )\right]=1 \Rightarrow f ^{\prime}[ g (2)] g ^{\prime}(2)=12 \Rightarrow f ^{\prime}( a ) g ^{\prime}(2) [\text { putting } x =2] $
$\text { given, } f ^{\prime}( a )=\frac{ a ^{10}}{1+ a ^2} \Rightarrow g ^{\prime}(2)=\frac{1+ a ^2}{ a ^{10}}$
$\text { Altemative: } g[f(x)]=x $
$g ^{\prime}[ f ( x )] \cdot f ^{\prime}( x )=1$
$\text { now } g (2)= a \Rightarrow f ( a )=2 $
$\therefore g \text { and } f \text { are inverse of each other } $
$\text { now } f ( x )=2 \Rightarrow g (2)= x = a $
$\therefore g^{\prime}(2) \cdot f^{\prime}(a)=1 $
$g^{\prime}(2)=\frac{1}{f^{\prime}(a)}=\frac{1+a^2}{a^{10}}$
