If `y=log_(e)x+sinx+e^(x)" then "(dy)/(dx)` is

Options

A. `(1)/(x)+sinx+e^(x)`
B. `(1)/(x)-cosx+e^(x)`
C. `(1)/(x)+cosx+e^(x)`
D. `(1)/(x)-sinx`

Answer

To find the derivative of the function \( y = \log_e x + \sin x + e^x \), we will differentiate each term separately with respect to \( x \).

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = \log_e x + \sin x + e^x \]

2. **Differentiate each term**:
- The derivative of \( \log_e x \) (which is also written as \( \ln x \)) is:
\[ \frac{d}{dx}(\log_e x) = \frac{1}{x} \]
- The derivative of \( \sin x \) is:
\[ \frac{d}{dx}(\sin x) = \cos x \]
- The derivative of \( e^x \) is:
\[ \frac{d}{dx}(e^x) = e^x \]

3. **Combine the derivatives**:
Now, we can combine the derivatives we calculated:
\[ \frac{dy}{dx} = \frac{1}{x} + \cos x + e^x \]

4. **Final result**:
Therefore, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{1}{x} + \cos x + e^x \]

### Conclusion:
The correct answer is:
\[ \frac{dy}{dx} = \frac{1}{x} + \cos x + e^x \]