Question

Let $g$ be the inverse function of $f$ and $f^{\prime }\left(x\right)=\frac{x^{10}}{1+x^2}.$ If $g\left(2\right)=a,$ then $g^{\prime }\left(2\right)$ is equal to

• A. $\frac{5}{2^{10}}$
• B. $\frac{1+a^2}{a^{10}}$
• C. $\frac{a^{10}}{1+a^2}$
• D. $\frac{1+a^{10}}{a^2}$

Answer 1

The correct option is B $\frac{1+a^2}{a^{10}}$

Given $g$ is inverse of $f$
So, $f\left(g\right(x\left)\right)=x$
$\Rightarrow f^{\prime }\left(g\right(x\left)\right)\cdot g^{\prime }\left(x\right)=1$
$\Rightarrow f^{\prime }\left(g\right(2\left)\right)\cdot g^{\prime }\left(2\right)=1$ (Putting $x=2$)
$\Rightarrow f^{\prime }\left(a\right)\cdot g^{\prime }\left(2\right)=1$

Given, $f^{\prime }\left(a\right)=\frac{a^{10}}{1+a^2}$
$\Rightarrow g^{\prime }\left(2\right)=\frac{1+a^2}{a^{10}}$