Question

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

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

Answer 1

Answer: (B)

$\frac{1+a^2}{a^{10}}$

Given that $g$ is the inverse function of $f$

Therefore, $f\left(g\right(x\left)\right)=x$

differentiating the above equation we get

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

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

Therefore, $f^{\prime }\left(g\right(x\left)\right)=\frac{\left(g\right(x){)}^{10}}{1+\left(g\right(x){)}^2}$

$⟹f^{\prime }\left(g\right(2\left)\right)=\frac{\left(g\right(2){)}^{10}}{1+\left(g\right(2){)}^2}$

$⟹f^{\prime }\left(g\right(2\left)\right)=\frac{a^{10}}{1+a^2}$ ------(2)

for $x=2$, eqn (1) becomes

$f^{\prime }\left(g\right(2\left)\right)\ast g^{\prime }\left(2\right)=1$ ------(3)

substituting (2) in (3) we get

$\frac{a^{10}}{1+a^2}\ast g^{\prime }\left(2\right)=1$

$⟹g^{\prime }\left(2\right)=\frac{1+a^2}{a^{10}}$