Question

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

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

Answer 1

Answer: (A)

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