Differentiate `tan^(-1)(secx+tanx),(-pi)/(2) lt x lt (pi)/(2)`

Explanation

To differentiate the function \( y = \tan^{-1}(\sec x + \tan x) \), we will follow these steps:

### Step 1: Differentiate the function
We start with the function:
\[ y = \tan^{-1}(\sec x + \tan x) \]
To differentiate \( y \) with respect to \( x \), we apply the chain rule:
\[ \frac{dy}{dx} = \frac{1}{1 + (\sec x + \tan x)^2} \cdot \frac{d}{dx}(\sec x + \tan x) \]

### Step 2: Differentiate \( \sec x + \tan x \)
Next, we need to differentiate \( \sec x + \tan x \):
\[ \frac{d}{dx}(\sec x + \tan x) = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) \]
Using the derivatives:
\[ \frac{d}{dx}(\sec x) = \sec x \tan x \quad \text{and} \quad \frac{d}{dx}(\tan x) = \sec^2 x \]
Thus,
\[ \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x \]

### Step 3: Substitute back into the derivative
Now we substitute this result back into our derivative:
\[ \frac{dy}{dx} = \frac{1}{1 + (\sec x + \tan x)^2} \cdot (\sec x \tan x + \sec^2 x) \]

### Step 4: Simplify the expression
We know that:
\[ 1 + \tan^2 x = \sec^2 x \]
Thus, we can express \( 1 + (\sec x + \tan x)^2 \) as:
\[ 1 + (\sec x + \tan x)^2 = 1 + \sec^2 x + 2\sec x \tan x + \tan^2 x = 1 + \sec^2 x + \tan^2 x + 2\sec x \tan x \]
Using \( 1 + \tan^2 x = \sec^2 x \):
\[ = \sec^2 x + \sec^2 x + 2\sec x \tan x = 2\sec^2 x + 2\sec x \tan x \]
Thus, we can write:
\[ \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{2\sec^2 x + 2\sec x \tan x} \]

### Step 5: Factor out common terms
Factoring out the common terms in the denominator:
\[ \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{2(\sec^2 x + \sec x \tan x)} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{1}{2} \]

### Final Result
Thus, the derivative of \( y = \tan^{-1}(\sec x + \tan x) \) is:
\[ \frac{dy}{dx} = \frac{1}{2} \]