Q. If $f(x) = |\log_{e}{|x|}|$, then $f'(x)$ equals
A) $\frac{1}{|x|},\ x \neq 0$
B) $\frac{1}{x}$ for $|x| > 1$ and $\frac{- 1}{x}$ for $|x| < 1$
C) $\frac{- 1}{x}$ for $|x| > 1$ and $\frac{1}{x}$ for $|x| < 1$
D) $\frac{1}{x}$ for $|x| > 0$ and $- \frac{1}{x}$ for $x < 0$
Explanation
(b)
For $x > 1$, we have
$f(x) = |\log{|x|| = \log{x\ \ \ \ \ \Rightarrow \ \ \ \ f^{'}(x) = \frac{1}{x}}}$
For $x < - 1$, we have
$f(x) = |\log{|x|| = \log{( - x)\ \ \ \ \ \Rightarrow \ \ \ \ f^{'}(x) = \frac{1}{x}}}$
For $0 < x < 1$, we have
$f(x) = |\log{|x|| = - \log{x\ \ \ \ \ \Rightarrow \ \ \ \ f^{'}(x) = \frac{- 1}{x}}}$
For $- 1 < x < 0$, we have
$f(x) = - {\log{( - x)\ \ \ \ \Rightarrow \ \ \ \ f^{'}(x) = - \frac{1}{x}}}$
Hence, $f^{'}(x) = \left\{ \begin{array}{r}
\frac{1}{x},\ \ |x| > 1 \\
- \frac{1}{x},\ \ \ |x| < 1
\end{array} \right.\$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.