Differentiate the following w.r.t. x :
`(logx)/e^(x)`

Answer

To differentiate the function \( y = \frac{\log x}{e^x} \) with respect to \( x \), we will use the quotient rule. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then the derivative is given by:

\[ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]

where \( u = \log x \) and \( v = e^x \).

### Step-by-step solution:

1. **Identify \( u \) and \( v \)**:
- Let \( u = \log x \)
- Let \( v = e^x \)

2. **Find \( \frac{du}{dx} \) and \( \frac{dv}{dx} \)**:
- The derivative of \( u \):
\[ \frac{du}{dx} = \frac{1}{x} \]
- The derivative of \( v \):
\[ \frac{dv}{dx} = e^x \]

3. **Apply the quotient rule**:
- Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \) into the quotient rule formula:
\[ \frac{dy}{dx} = \frac{e^x \cdot \frac{1}{x} - \log x \cdot e^x}{(e^x)^2} \]

4. **Simplify the expression**:
- Factor out \( e^x \) from the numerator:
\[ \frac{dy}{dx} = \frac{e^x \left( \frac{1}{x} - \log x \right)}{e^{2x}} \]
- This simplifies to:
\[ \frac{dy}{dx} = \frac{1 - x \log x}{x e^x} \]

5. **Final result**:
- Thus, the derivative of \( y = \frac{\log x}{e^x} \) is:
\[ \frac{dy}{dx} = \frac{1 - x \log x}{x e^x} \]