If `y = tan^(-1) (sec x + tan x) " then " (dy)/(dx)=` ?

Options

A. `(1)/(2)`
B. `(-1)/(2)`
C. 1
D. none of these

Explanation

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

### Step 1: Differentiate the function
We start with the function:
\[ y = \tan^{-1}(\sec x + \tan x) \]
Using the chain rule for differentiation, we have:
\[ \frac{dy}{dx} = \frac{1}{1 + (\sec x + \tan x)^2} \cdot \frac{d}{dx}(\sec x + \tan x) \]

### Step 2: Differentiate \( \sec x + \tan x \)
Now we need to differentiate \( \sec x + \tan x \):
\[ \frac{d}{dx}(\sec x) = \sec x \tan x \]
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]
Thus, we have:
\[ \frac{d}{dx}(\sec x + \tan x) = \sec x \tan x + \sec^2 x \]

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

### Step 4: Simplify the expression
Now we need to simplify \( 1 + (\sec x + \tan x)^2 \):
\[ (\sec x + \tan x)^2 = \sec^2 x + 2\sec x \tan x + \tan^2 x \]
Using the identity \( \sec^2 x - \tan^2 x = 1 \), we can rewrite:
\[ \sec^2 x + \tan^2 x = 1 + 2\sec x \tan x \]
Thus:
\[ 1 + (\sec x + \tan x)^2 = 1 + (1 + 2\sec x \tan x) = 2 + 2\sec x \tan x \]

### Step 5: Final expression for the derivative
Now substituting this back, we get:
\[ \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{2 + 2\sec x \tan x} \]
We can factor out a 2 from the denominator:
\[ \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{2(1 + \sec x \tan x)} \]

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

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