Question

The function $f(x)=\underset{4}{\overset{x^2}{\int }}\sqrt{9+t^2}dt$ has an inverse. The value of ${\left(f^{-1}\right)}^{\prime }(0)$ equal

• A. $\frac{1}{10}$
• B. $\frac{1}{20}$
• C. $\frac{1}{4\sqrt{13}}$
• D. $\frac{1}{5}$

Answer 1

Answer: (B)

$y=0,x=2($ we have to find $\frac{dx}{dy}$ when $y=0$ )
$f^{\prime }(x)=\sqrt{9+x^4}\cdot 2x$
$\therefore g^{\prime }(y)={\frac{1}{f^{\prime }(x)}|}_{x=2}=\frac{1}{2x\sqrt{9+x^4}}=\frac{1}{20}$