Question

Let $f$ be a one-one function satisfying $f^{\prime }\left(x\right)=f\left(x\right)$ then ${\left(f^{-1}\right)}^{\prime \prime }\left(x\right)$ 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)

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

Let $f$ be the one-one function

Given : since $f^1\left(x\right)=f\left(x\right)\Rightarrow f\left(x\right)-f^1\left(x\right)=0$

$\frac{df\left(x\right)}{dx}=f\left(x\right)$

Integrating both left and right handside

$\int \frac{df\left(x\right)}{f\left(x\right)}=\int dx$

$\ln \left[f\left(x\right)\right]=x+c$

$f\left(x\right)=e^x+c$

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

$f^{-1}\left(e^x\right)=\frac{e^x}{f}$ ...Differntiate w.r.t x

$=\frac{e^x\left(f\right)-f^1\left(e^x\right)}{f^2}$

$\frac{e^x(f-f^1)}{f^{}}$

$=0$

${\left(f^1\right)}^{{}^{\prime \prime }}\left(e^x\right)=-\left[\frac{1(2-1)-0}{x^2}\right]$

${\left(f^{-1}\right)}^4\left(x\right)=-\frac{1}{x^2}$

Therefore option $B$ is a correct answer