Question

Let $\phi(x)$ be the inverse of the function $f(x)$ and $f^{\prime} x=\frac{1}{1+x^{5}}$, then $\frac{d}{d x} \phi x$ is equal to

• A. $\frac{1}{1+\phi\, x^{5}}$
• B. $\frac{1}{1+f x^{5}}$
• C. $1+\phi\, x^{5}$
• D. $1+f(x)$

Answer 1

Answer: (C)

We have, $\phi x=f^{-1}(x)$
$\Rightarrow x=f[\phi x]$
On differentiating both sides w.r.t. $x$, we get
$1=f^{\prime} \phi x \cdot \phi^{\prime}(x) $
$\Rightarrow \phi^{\prime} x=\frac{1}{f^{\prime} \phi x} \ldots \text { (i) }$
Since, $f^{\prime} x=\frac{1}{1+x^{5}}$ (given)
$f^{\prime} \phi x=\frac{1}{1+\phi x^{5}}$
From Equation (i),
$\phi^{\prime} x=\frac{1}{f^{\prime}[\phi x]}=1+\phi x^{5}$