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Q.
The function $f( x )=\int\limits_4^{ x ^2} \sqrt{9+ t ^2} dt$ has an inverse. The value of $\left(f^{-1}\right)^{\prime}(0)$ equal
Integrals
Solution:
$y=0, x=2\left(\right.$ we have to find $\frac{ dx }{ dy }$ when $y =0$ )
$f^{\prime}(x)=\sqrt{9+x^4} \cdot 2 x $
$\therefore g^{\prime}(y)=\left.\frac{1}{f^{\prime}(x)}\right|_{x=2}=\frac{1}{2 x \sqrt{9+x^4}}=\frac{1}{20}$