Question

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

• A. $1+\left[g\right(x){]}^n$
• B. $1-g\left(x\right)$
• C. $1+g\left(x\right)$
• D. $-g(x{)}^n$

Answer 1

Answer: (A)

$1+\left[g\right(x){]}^n$

g is inverse of f

$f^{\prime }\left(x\right)=\frac{1}{1+x^n}$

$\therefore$ $g\left(x\right)=f^{-1}\left(x\right)$

$f\circ g=I$ $[\because g(x)=f^{-1}(x\left)\right]$

$f\left(g\right(x\left)\right)=I\left(x\right)=x$

diff. both sides w.r.t $x$

$f^{\prime }\left(g\right(x\left)\right)$. $g^{\prime }\left(x\right)=1$

$\frac{1}{1+{\left[g\right(x\left)\right]}^n}$ $g^{\prime }\left(x\right)=1$

$g^{\prime }\left(x\right)=1+{\left[g\right(x\left)\right]}^n$