A) \[\frac{1}{x}\]
B) \[-\frac{1}{x}\]
C) \[\pm \,\frac{1}{2x}\]
D) None of these
Correct Answer: A
Solution :
If \[x\ne 0,\,\,y\,\,{{\log }_{e}}\,|2x|\] \[y=\left\{ \begin{matrix} {{\log }_{e}}(-2x), & if\,x<0 \\ {{\log }_{e}}\,(2x), & if\,x>0 \\ \end{matrix} \right.\] \[\frac{dy}{dx}=y'=\left\{ \begin{matrix} \frac{1}{(-2x)}(-2), & if\,x<0 \\ \frac{1}{(2x)}\,(2), & if\,x>0 \\ \end{matrix} \right.\] \[=\left\{ \begin{matrix} 1/x, & if\,x<0 \\ 1/x, & if\,x>0 \\ \end{matrix} \right.\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{x},\,\,\,if\,\,x\ne 0\]
You need to login to perform this action.
You will be redirected in
3 sec
