If `f(x)=log|2x|,xne0` then `f'(x)` is equal to

Options

A. `(1)/(x)`
B. `-(1)/(x)`
C. `(1)/(|x|)`
D. none of these

Explanation

To find the derivative of the function \( f(x) = \log |2x| \) for \( x \neq 0 \), we will follow these steps:

### Step 1: Rewrite the function based on the sign of \( x \)
The absolute value function \( |2x| \) can be expressed differently depending on whether \( x \) is positive or negative:
- If \( x > 0 \), then \( |2x| = 2x \).
- If \( x < 0 \), then \( |2x| = -2x \).

Thus, we can rewrite \( f(x) \) as:
\[ f(x) = \begin{cases} \log(2x) & \text{if } x > 0 \\ \log(-2x) & \text{if } x < 0 \end{cases} \]

### Step 2: Differentiate \( f(x) \) for \( x > 0 \)
Using the chain rule, the derivative of \( f(x) = \log(2x) \) is:
\[ f'(x) = \frac{d}{dx} \log(2x) = \frac{1}{2x} \cdot \frac{d}{dx}(2x) = \frac{1}{2x} \cdot 2 = \frac{1}{x} \]

### Step 3: Differentiate \( f(x) \) for \( x < 0 \)
Similarly, for \( f(x) = \log(-2x) \):
\[ f'(x) = \frac{d}{dx} \log(-2x) = \frac{1}{-2x} \cdot \frac{d}{dx}(-2x) = \frac{1}{-2x} \cdot (-2) = \frac{1}{x} \]

### Step 4: Combine the results
From both cases, we find that:
\[ f'(x) = \frac{1}{x} \quad \text{for both } x > 0 \text{ and } x < 0 \]

### Final Answer
Thus, the derivative \( f'(x) \) is:
\[ f'(x) = \frac{1}{x} \quad \text{for all } x \neq 0 \]

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