Question

If $f\left(x\right)={\log }_e\left\{\frac{u\left(x\right)}{v\left(x\right)}\right\},u\left(1\right)=v\left(1\right)$ and $u^{\prime }\left(x\right)=v^{\prime }\left(x\right)=2$, then $f^{\prime }\left(1\right)$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. none of these

Answer 1

Answer: (A)

$0$

$f\left(x\right)={\log }_e\left\{\frac{u\left(x\right)}{v\left(x\right)}\right\},u\left(1\right)=v\left(1\right)$ and $u^{\prime }\left(x\right)=v^{\prime }\left(x\right)=2$
$f^{\prime }\left(x\right)=\frac{d}{dx}\left({\log }_e\frac{u\left(x\right)}{v\left(x\right)}\right)$
$\Rightarrow f^{\prime }\left(x\right)=\frac{1}{\frac{u\left(x\right)}{v\left(x\right)}}\frac{d}{dx}\left[\frac{u\left(x\right)}{v\left(x\right)}\right]$
$\Rightarrow f^{\prime }\left(x\right)=\frac{v\left(x\right)}{u\left(x\right)}\left[\frac{v\left(x\right).u^{\prime }\left(x\right)-u\left(x\right).v^{\prime }\left(x\right)}{{\left(v\left(x\right)\right)}^2}\right]$
$f^{\prime }\left(1\right)=\frac{v\left(1\right).u^{\prime }\left(1\right)-u\left(1\right).v^{\prime }\left(1\right)}{v\left(1\right)}=\frac{v\left(1\right).2-v\left(1\right).2}{v\left(1\right)}=0$