Differentiate the following w.r.t. x :
`ln(secx+tanx)`

Answer

To differentiate the function \( y = \ln(\sec x + \tan x) \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate the logarithmic function
Using the chain rule, the derivative of \( y = \ln(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = \sec x + \tan x \).

### Step 2: Find \( \frac{du}{dx} \)
Now we need to differentiate \( u = \sec x + \tan x \):
- The derivative of \( \sec x \) is \( \sec x \tan x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).

Thus, we have:
\[ \frac{du}{dx} = \sec x \tan x + \sec^2 x \]

### Step 3: Substitute back into the derivative formula
Now substituting \( u \) and \( \frac{du}{dx} \) back into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{\sec x + \tan x} \cdot \left( \sec x \tan x + \sec^2 x \right) \]

### Step 4: Simplify the expression
We can simplify the expression:
\[ \frac{dy}{dx} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} \]

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