Question

If $g\left(x\right)$ is the inverse function of $f\left(x\right)$ and $f^{\prime }\left(x\right)=\frac{1}{1+x^4}$, then $g^{\prime }\left(x\right)$ is

• A. $1+\left[g\right(x){]}^4$
• B. $1-\left[g\right(x){]}^4$
• C. $1+\left[f\right(x){]}^4$
• D. $\frac{1}{1+\left[g\right(x){]}^4}$

Answer 1

The correct option is A $1+\left[g\right(x){]}^4$

$g=f^{-1}$
$f\left(g\right(x\left)\right)=x$
Differentiate $w.r.t.x$
$f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)=1$
$\therefore \frac{1}{1+\left(g\right(x){)}^4}\cdot g^{\prime }\left(x\right)=1$
$g^{\prime }\left(x\right)=1+\left[g\right(x){]}^4$