Let $\phi(x)$ be the inverse of the function f(x) and $f'(x)=\frac{1}{1+x^5}$, then $\frac{d}{dx}\phi(x)$ is
Answer & explanation
Correct answer: option 3
$\phi(x)=f^{-1}(x)$
$⇒x=f(\phi(x))⇒1=(f'(\phi(x))).\phi'(x)$
$⇒\phi'(x)=\frac{1}{f'(\phi(x))}⇒f'(x)=\frac{1}{1+x^5}⇒f'(\phi(x))=\frac{1}{1+(\phi(x))^5}$
or $\phi'(x)=\frac{1}{f'(\phi(x))}=1+(\phi(x))^5$