Question

Suppose $f$ and $g$ are functions having second derivative $f\text{'}\text{'}$ and $g\text{'}\text{'}$ respectively at all points. If $f\left(x\right)\cdot g\left(x\right)=1$ for all $x$ and $f\text{'}$ and $g\text{'}$ are never zero, then $\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\left(x\right)}-\frac{g\text{'}\text{'}\left(x\right)}{g\text{'}\left(x\right)}$equals

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

Answer 1

Answer: (B), (D)

$\frac{2f^{\prime }\left(x\right)}{f\left(x\right)}$

We know that,
$f\left(x\right)\cdot g\left(x\right)=1\Rightarrow g=\frac{1}{f}\Rightarrow g^{\prime }=-\frac{1}{f^2}{f}^{\prime }\Rightarrow g^{\prime \prime }=-[-\frac{2}{f^3}{f}^{\prime 2}+\frac{1}{f^2}{f}^{\prime \prime }]$
So,
$\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\left(x\right)}-\frac{g\text{'}\text{'}\left(x\right)}{g\text{'}\left(x\right)}=\frac{f\text{'}\text{'}}{f\text{'}}-\frac{-[-\frac{2}{f^3}{f}^{\prime 2}+\frac{1}{f^2}{f}^{\prime \prime }]}{-\frac{1}{f^2}{f}^{\prime }}=\frac{f\text{'}\text{'}}{f\text{'}}-(\frac{-2f^{\prime }}{f}+\frac{f^{\prime \prime }}{f})=\frac{2f^{\prime }}{f}$

Also,
$g\cdot f=1\Rightarrow g^{\prime }f+g{f}^{\prime }=0\Rightarrow \frac{f^{\prime }}{f}=-\frac{g^{\prime }}{g}$

Hence,
$\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\left(x\right)}-\frac{g\text{'}\text{'}\left(x\right)}{g\text{'}\left(x\right)}=\frac{2f^{\prime }\left(x\right)}{f\left(x\right)}=-\frac{2g^{\prime }\left(x\right)}{g\left(x\right)}$