Question

Let $f$ be a bijection satisfying $f^{\prime }\left(x\right)=f\left(x\right)$. Then, $\left(f^{-1}{)}^{\prime \prime }\right(x)$ is equal to

• A. $-\frac{1}{x^3}$
• B. $-\frac{1}{x^2}$
• C. $f\left(x\right)$
• D. $f^{-1}\left(x\right)$

Answer 1

Answer: (B)

$-\frac{1}{x^2}$

We have,
$f^{-1}\left(f\right(x\left)\right)=x$ for all $x$
$\Rightarrow {\left(f^{-1}\right)}^{\prime }\left(f\right(x\left)\right)f^{\prime }\left(x\right)=1$ for all $x$
$\Rightarrow {\left(f^{-1}\right)}^{\prime }\left(f\right(x\left)\right)=\frac{1}{f\left(x\right)}$ for all $x[\because f^{\prime }(x)=f(x\left)\right]$
$\Rightarrow {\left(f^{-1}\right)}^{\prime \prime }\left(f\right(x\left)\right)f^{\prime }\left(x\right)=-\frac{1}{|f\left(x\right){|}^2}{f}^{\prime }\left(x\right)[$ Differentiating both sides w.r.t. $x]$
$\Rightarrow {\left(f^{-1}\right)}^{\prime \prime }\left(f\right(x\left)\right)=-\frac{1}{|f\left(x\right){|}^2}$
$\Rightarrow {\left(f^{-1}\right)}^{\prime \prime }\left(x\right)=-\frac{1}{x^2}$