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Q.
Let $f ( x )=\int\limits_2^{ x } \frac{ dt }{\sqrt{1+ t ^4}}$ and $g$ be the inverse of $f$. Then the value of $g ^{\prime}(0)$ is -
Integrals
Solution:
$f ( x )=\int\limits_2^{ x } \frac{ dt }{\sqrt{1+ t ^4}} [\because f (2)=0]$
$g ( f ( x ))= x$
$g ^{\prime}( f ( x )) f ^{\prime}( x )=1$
$g^{\prime}(0) f^{\prime}(2)=2$
$g^{\prime}(0)=\frac{1}{f^{\prime}(2)}=\sqrt{17}$
$\Rightarrow f^{\prime}(x)=\frac{1}{\sqrt{1+x^4}} \Rightarrow f^{\prime}(2)=\frac{1}{\sqrt{17}}$