Question

If $x\ne 0$and $y={\log }_e\left|2x\right|$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{1}{x}$
• B. $\frac{-1}{x}$
• C. $\pm \frac{1}{2x}$
• D. None of these

Answer 1

The correct option is A

$\frac{1}{x}$

Explanation for the correct option:

Finding the value of $\frac{dy}{dx}$:

Given,

$y={\log }_e\left|2x\right|$

Case 1:

If $x<0,y={\log }_e\left(-2x\right)$

On differentiating we get,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{-2x}\left(-2\right)\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]}\\ & \Rightarrow & \frac{dy}{dx}=\frac{1}{x}\end{array}$

Case 2:

If $x>0,y={\log }_e\left(2x\right)$

On differentiating we get,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{1}{2x}\left(2\right)\mathbf{[}\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{l}\mathbf{o}\mathbf{g}\mathbf{}\mathbf{x}\mathbf{=}\frac{\mathbf{1}}{\mathbf{x}}\mathbf{]}\\ & \Rightarrow & \frac{dy}{dx}=\frac{1}{x}\end{array}$

From the above two cases $\frac{dy}{dx}=\frac{1}{x}$.

Hence, the correct option is A.