`y=e^(x)lnx`

Options

A. `e^(x)lnx-(e^(x))/(x)`
B. `e^(x)lnx-(2e^(x))/(x)`
C. `2e^(x)lnx-(e^(x))/(x)`
D. `e^(x)lnx+(e^(x))/(x)`

Answer

To differentiate the function \( y = e^x \ln x \), we will use the product rule of differentiation. The product rule states that if you have two functions \( u(x) \) and \( v(x) \), the derivative of their product is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the Functions**:
- Let \( u = e^x \)
- Let \( v = \ln x \)

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

3. **Apply the Product Rule**:
- Now, using the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
- Substitute \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = e^x \cdot \frac{1}{x} + \ln x \cdot e^x \]

4. **Simplify the Expression**:
- Combine the terms:
\[ \frac{dy}{dx} = \frac{e^x}{x} + e^x \ln x \]
- Factor out \( e^x \):
\[ \frac{dy}{dx} = e^x \left( \frac{1}{x} + \ln x \right) \]

5. **Final Result**:
- The derivative of \( y = e^x \ln x \) is:
\[ \frac{dy}{dx} = e^x \left( \frac{1}{x} + \ln x \right) \]