Differentiate : tan^(-1)(secx+tanx)

Explanation

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

### Step 1: Rewrite the function
We start with the function:
\[ y = \tan^{-1}(\sec x + \tan x) \]

### Step 2: Use trigonometric identities
Recall the identities:
- \( \sec x = \frac{1}{\cos x} \)
- \( \tan x = \frac{\sin x}{\cos x} \)

Thus, we can rewrite \( \sec x + \tan x \):
\[ \sec x + \tan x = \frac{1}{\cos x} + \frac{\sin x}{\cos x} = \frac{1 + \sin x}{\cos x} \]

Now, substituting this back into our function:
\[ y = \tan^{-1}\left(\frac{1 + \sin x}{\cos x}\right) \]

### Step 3: Differentiate using the chain rule
To differentiate \( y \), we apply the chain rule:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{1 + \sin x}{\cos x}\right)^2} \cdot \frac{d}{dx}\left(\frac{1 + \sin x}{\cos x}\right) \]

### Step 4: Differentiate the inner function
Now, we differentiate the inner function \( \frac{1 + \sin x}{\cos x} \) using the quotient rule:
\[ \frac{d}{dx}\left(\frac{1 + \sin x}{\cos x}\right) = \frac{\cos x \cdot \cos x - (1 + \sin x)(-\sin x)}{\cos^2 x} \]
This simplifies to:
\[ \frac{\cos^2 x + (1 + \sin x)\sin x}{\cos^2 x} = \frac{\cos^2 x + \sin x + \sin^2 x}{\cos^2 x} = \frac{1 + \sin x}{\cos^2 x} \]

### Step 5: Substitute back into the derivative
Substituting this back into our derivative:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{1 + \sin x}{\cos x}\right)^2} \cdot \frac{1 + \sin x}{\cos^2 x} \]

### Step 6: Simplify the expression
Now, we need to simplify \( 1 + \left(\frac{1 + \sin x}{\cos x}\right)^2 \):
\[ 1 + \left(\frac{1 + \sin x}{\cos x}\right)^2 = 1 + \frac{(1 + \sin x)^2}{\cos^2 x} = \frac{\cos^2 x + (1 + \sin x)^2}{\cos^2 x} \]
Expanding \( (1 + \sin x)^2 \):
\[ (1 + \sin x)^2 = 1 + 2\sin x + \sin^2 x \]
Thus:
\[ 1 + \left(\frac{1 + \sin x}{\cos x}\right)^2 = \frac{\cos^2 x + 1 + 2\sin x + \sin^2 x}{\cos^2 x} = \frac{2 + 2\sin x}{\cos^2 x} = \frac{2(1 + \sin x)}{\cos^2 x} \]

### Final expression for the derivative
Substituting this back into the derivative:
\[ \frac{dy}{dx} = \frac{1 + \sin x}{\cos^2 x} \cdot \frac{\cos^2 x}{2(1 + \sin x)} = \frac{1}{2} \]

Thus, the final answer is:
\[ \frac{dy}{dx} = \frac{1}{2} \]