If f(x) =| log x|, then `

Options

A. `y' (1+) =(1)/(x)`
B. `y'(1-) =-(1)/(x)`
C. `y'(1)=1`
D. `y'(0) =oo`

Answer

To find the derivative of the function \( f(x) = |\log x| \), we need to analyze the function based on the properties of the logarithm and the absolute value.

### Step-by-Step Solution:

1. **Identify the function**:
The function is given as \( f(x) = |\log x| \).

2. **Determine the behavior of \( \log x \)**:
- For \( x > 1 \): \( \log x > 0 \) so \( |\log x| = \log x \).
- For \( 0 < x < 1 \): \( \log x < 0 \) so \( |\log x| = -\log x \).
- At \( x = 1 \): \( \log 1 = 0 \) so \( f(1) = 0 \).

3. **Find the right-hand derivative (RHD) at \( x = 1 \)**:
\[ f'(1^+) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} \]
For \( 1+h > 1 \) (i.e., \( h > 0 \)):
\[ f(1+h) = \log(1+h) \]
Therefore,
\[ f'(1^+) = \lim_{h \to 0^+} \frac{\log(1+h) - 0}{h} = \lim_{h \to 0^+} \frac{\log(1+h)}{h} \]
Using the expansion \( \log(1+h) \approx h \) as \( h \to 0 \):
\[ f'(1^+) = \lim_{h \to 0^+} \frac{h}{h} = 1 \]

4. **Find the left-hand derivative (LHD) at \( x = 1 \)**:
\[ f'(1^-) = \lim_{h \to 0^+} \frac{f(1-h) - f(1)}{-h} \]
For \( 1-h < 1 \) (i.e., \( h > 0 \)):
\[ f(1-h) = -\log(1-h) \]
Therefore,
\[ f'(1^-) = \lim_{h \to 0^+} \frac{-\log(1-h)}{-h} = \lim_{h \to 0^+} \frac{\log(1-h)}{h} \]
Using the expansion \( \log(1-h) \approx -h \) as \( h \to 0 \):
\[ f'(1^-) = \lim_{h \to 0^+} \frac{-h}{h} = -1 \]

5. **Conclusion about the derivative at \( x = 1 \)**:
Since \( f'(1^+) = 1 \) and \( f'(1^-) = -1 \), the left-hand and right-hand derivatives at \( x = 1 \) are not equal. Therefore, the derivative \( f'(1) \) is not defined.

6. **Evaluate \( f'(x) \) for \( x < 1 \) and \( x > 1 \)**:
- For \( x > 1 \):
\[ f'(x) = \frac{1}{x} \]
- For \( 0 < x < 1 \):
\[ f'(x) = -\frac{1}{x} \]

### Final Result:
- \( f'(x) = \frac{1}{x} \) for \( x > 1 \)
- \( f'(x) = -\frac{1}{x} \) for \( 0 < x < 1 \)
- \( f'(1) \) is not defined.