`d/(dx)log_|x|e=`

Options

A. `e^x`
B. `1/(logx)^2`
C. `(-1)/(x(log|x|)^2)`
D. `1/|x|`

Answer

To solve the differentiation problem \( \frac{d}{dx} \log_{|x|} e \), we can follow these steps:

### Step-by-Step Solution:

1. **Rewrite the logarithm**:
We start by rewriting the logarithm using the change of base formula:
\[ \log_{|x|} e = \frac{\log_e e}{\log_e |x|} = \frac{1}{\log_e |x|} \]
Here, \( \log_e e = 1 \).

2. **Differentiate the expression**:
Now, we differentiate \( y = \frac{1}{\log_e |x|} \) using the quotient rule or the chain rule. We can rewrite it as:
\[ y = (\log_e |x|)^{-1} \]
Using the chain rule, we have:
\[ \frac{dy}{dx} = -1 \cdot (\log_e |x|)^{-2} \cdot \frac{d}{dx}(\log_e |x|) \]

3. **Differentiate \( \log_e |x| \)**:
The derivative of \( \log_e |x| \) is:
\[ \frac{d}{dx}(\log_e |x|) = \frac{1}{|x|} \cdot \frac{d}{dx}(|x|) = \frac{1}{|x|} \cdot \text{sgn}(x) \]
where \( \text{sgn}(x) \) is the sign function, which is \( 1 \) for \( x > 0 \) and \( -1 \) for \( x < 0 \).

4. **Combine the results**:
Substituting back into our derivative, we get:
\[ \frac{dy}{dx} = -\frac{1}{(\log_e |x|)^2} \cdot \frac{1}{|x|} \cdot \text{sgn}(x) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\text{sgn}(x)}{|x| (\log_e |x|)^2} \]

5. **Final result**:
Thus, the derivative of \( \log_{|x|} e \) with respect to \( x \) is:
\[ \frac{d}{dx} \log_{|x|} e = -\frac{\text{sgn}(x)}{|x| (\log_e |x|)^2} \]