Question

If the function $f\left(x\right)=x^5+e^{\frac{x}{5}}$ and $g\left(x\right)=f^{-1}\left(x\right)$, then the value of $\frac{1}{g^{{}^{\prime }}(1+e^{\frac{1}{5}})}$ is -

• A. $5$
• B. $5+\frac{e^{\frac{1}{5}}}{5}$
• C. $1$
• D. $5+\frac{5}{e}$

Answer 1

The correct option is B $5+\frac{e^{\frac{1}{5}}}{5}$

$fo{f}^{-1}\left(x\right)=x\Rightarrow fog\left(x\right)=x$
On differentiating both sides, we get-
$f^{{}^{\prime }}\left(g\right(x\left)\right)\cdot g^{{}^{\prime }}\left(x\right)=1$, where $f^{{}^{\prime }}\left(x\right)=5x^4+\frac{1}{5}{e}^{\frac{x}{5}}$
$g^{{}^{\prime }}\left(x\right)=\frac{1}{f^{{}^{\prime }}\left(g\right(x\left)\right)}$
$g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(g\right(f\left(x\right)\left)\right)}$ [Replace $x$ by $f\left(x\right)$]
$g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}$ [Since, $g\left(f\right(x\left)\right)=fog\left(x\right)=x$]
$\Rightarrow g^{{}^{\prime }}(1+e^{\frac{1}{5}})=\frac{1}{f^{{}^{\prime }}\left(1\right)}=\frac{1}{5+\frac{1}{5}{e}^{\frac{1}{5}}}$