Question

If $g$ is the inverse of $f$ and $f^{\prime }\left(x\right)=\frac{1}{2+x^n}$ , then $g^{\prime }\left(x\right)$

• A. $2+x^n$
• B. $2+\left(f\right(x){)}^n$
• C. $2+\left(g\right(x){)}^n$
• D. $0$

Answer 1

Answer: (C)

$2+\left(g\right(x){)}^n$

Given $f\left(x\right)=\frac{1}{2+x^n}$
$y=\frac{1}{2+x^n}$
$\Rightarrow 2+x^n=\frac{1}{y}$
$\Rightarrow x^n=\frac{1-2y}{y}$
$\Rightarrow x={\left(\frac{1-2y}{y}\right)}^{1/n}$
$\Rightarrow f^{-1}\left(y\right)={\left(\frac{1-2y}{y}\right)}^{1/n}$
$\Rightarrow g\left(x\right)=f^{-1}\left(x\right)={\left(\frac{1-2x}{x}\right)}^{1/n}$
Now, $g^{\prime }\left(x\right)={\frac{-1}{x^{2}n}\left(\frac{1-2x}{x}\right)}^{(1-n)/n}$
$\Rightarrow g^{\prime }\left(x\right)=\frac{-1}{x^{2}n}\left(g\right(x){)}^{1-n}$