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 find \( f'(x) \). The function consists of three parts, so we will differentiate each part separately.

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

2. **Differentiate \( \ln(\sec x) \)**:
\[ \frac{d}{dx}(\ln(\sec x)) = \frac{1}{\sec x} \cdot \frac{d}{dx}(\sec x) = \frac{1}{\sec x} \cdot \sec x \tan x = \tan x \]
(Using the derivative of \( \sec x \), which is \( \sec x \tan x \)).

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

Combining these results, 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 need to evaluate \( f'(\frac{\pi}{4}) \):
1. **Calculate \( \tan(\frac{\pi}{4}) \)**:
\[ \tan\left(\frac{\pi}{4}\right) = 1 \]

2. **Calculate \( \sec^2(\frac{\pi}{4}) \)**:
\[ \sec\left(\frac{\pi}{4}\right) = \frac{1}{\cos\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \quad \Rightarrow \quad \sec^2\left(\frac{\pi}{4}\right) = 2 \]

3. **Substitute these values into \( f'(\frac{\pi}{4}) \)**:
\[ f'\left(\frac{\pi}{4}\right) = e^{\tan\left(\frac{\pi}{4}\right)} \sec^2\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right) - 1 \]
\[ = e^1 \cdot 2 + 1 - 1 \]
\[ = 2e + 1 - 1 = 2e \]

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

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