Differentiate the following w.r.t. x :
`tan^(-1)(secx+tanx)`.

Answer

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

### Step 1: Identify the function to differentiate
We have:
\[ y = \tan^{-1}(\sec x + \tan x) \]

### Step 2: Use the chain rule
To differentiate \( y \) with respect to \( x \), we apply the chain rule. The derivative of \( \tan^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \sec x + \tan x \).

### Step 3: Differentiate \( u = \sec x + \tan x \)
Now, we need to find \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) \]
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,
\[ \frac{du}{dx} = \sec x \tan x + \sec^2 x \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the chain rule formula
Now we substitute \( u \) and \( \frac{du}{dx} \) back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{1 + (\sec x + \tan x)^2} \cdot (\sec x \tan x + \sec^2 x) \]

### Step 5: Simplify \( 1 + (\sec x + \tan x)^2 \)
Next, 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 \( \sec^2 x = 1 + \tan^2 x \), we have:
\[ 1 + (\sec x + \tan x)^2 = 1 + \sec^2 x + 2\sec x \tan x + \tan^2 x = 2 + 2\sec x \tan x \]

### Step 6: Final expression for \( \frac{dy}{dx} \)
Now substituting this back, we get:
\[ \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{2 + 2\sec x \tan x} \]
This can be simplified further:
\[ \frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{2(1 + \sec x \tan x)} \]

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

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