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 equal to

Question

Dr. Abubacker · Accepted Answer

(c)
We have, $\phi(x) = f^{- 1}(x)$
$\Rightarrow x = f\lbrack\phi(x)\rbrack$
On differentiating both sides w.r.t. $x$, we get
$1 = f^{'}\left\lbrack \phi(x) \right\rbrack.\phi^{'}(x)$
$\Rightarrow \phi(x) = \frac{1}{f^{'}\lbrack\phi(x)\rbrack}$ …(i)
Since, $f^{'}(x) = \frac{1}{1 + x^{5}\ }$ (given)
$\Rightarrow f^{'}\lbrack\phi(x) = \frac{1}{1 + \left\lbrack \phi(x) \right\rbrack^{5}}$
From Eq.(i),
$\phi^{'}(x) = \frac{1}{f^{'}\lbrack\phi(x)\rbrack} = 1 + \left\lbrack \phi(x) \right\rbrack^{5}$

Q. 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 equal to

Explanation

Reviewed by Dr. Abubacker

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