What is the derivative of the function
`f(x)=e^(tanx)+ln(secx)-e^(lnx)" at "x=(pi)/(4)?`

Options

A. `(e)/(2)`
B. e
C. 2e
D. 4e

Answer

To find the derivative of the function \( f(x) = e^{\tan x} + \ln(\sec x) - e^{\ln x} \) at \( x = \frac{\pi}{4} \), we will follow these steps:

### Step 1: Differentiate the function
We need to differentiate each term in the function:

1. **Differentiate \( e^{\tan x} \)**:
- Using the chain rule, the derivative is:
\[ \frac{d}{dx}(e^{\tan x}) = e^{\tan x} \cdot \sec^2 x \]

2. **Differentiate \( \ln(\sec x) \)**:
- The derivative of \( \ln(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \sec x \):
\[ \frac{d}{dx}(\ln(\sec x)) = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x \]

3. **Differentiate \( -e^{\ln x} \)**:
- Since \( e^{\ln x} = x \), the derivative is:
\[ \frac{d}{dx}(-e^{\ln x}) = -1 \]

Putting it all together, we have:
\[ f'(x) = e^{\tan x} \sec^2 x + \tan x - 1 \]

### Step 2: Evaluate the derivative at \( x = \frac{\pi}{4} \)
Now we will substitute \( x = \frac{\pi}{4} \) into the derivative:

1. **Calculate \( e^{\tan(\frac{\pi}{4})} \)**:
- Since \( \tan(\frac{\pi}{4}) = 1 \):
\[ e^{\tan(\frac{\pi}{4})} = e^1 = e \]

2. **Calculate \( \sec^2(\frac{\pi}{4}) \)**:
- Since \( \sec(\frac{\pi}{4}) = \sqrt{2} \):
\[ \sec^2(\frac{\pi}{4}) = (\sqrt{2})^2 = 2 \]

3. **Calculate \( \tan(\frac{\pi}{4}) \)**:
- Again, \( \tan(\frac{\pi}{4}) = 1 \).

Putting these values into the derivative:
\[ f'\left(\frac{\pi}{4}\right) = e \cdot 2 + 1 - 1 = 2e \]

### Final Answer
Thus, the derivative of the function at \( x = \frac{\pi}{4} \) is:
\[ \boxed{2e} \]