derivative of log (sec 2x +tan 2 x)

Answer

To find the derivative of the function \( y = \log(\sec(2x) + \tan(2x)) \), we will follow these steps:

### Step 1: Identify the function
We start with the function:
\[ y = \log(\sec(2x) + \tan(2x)) \]

### Step 2: Apply the chain rule
Using the chain rule for differentiation, we know that the derivative of \( \log(u) \) is:
\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = \sec(2x) + \tan(2x) \).

### Step 3: Differentiate \( u \)
Now we need to find \( \frac{du}{dx} \):
\[ u = \sec(2x) + \tan(2x) \]
To differentiate \( u \), we will use the derivatives of \( \sec(x) \) and \( \tan(x) \):
\[ \frac{d}{dx}(\sec(x)) = \sec(x)\tan(x) \quad \text{and} \quad \frac{d}{dx}(\tan(x)) = \sec^2(x) \]
Using the chain rule for \( \sec(2x) \) and \( \tan(2x) \):
\[ \frac{du}{dx} = \frac{d}{dx}(\sec(2x)) + \frac{d}{dx}(\tan(2x)) = \sec(2x)\tan(2x) \cdot 2 + \sec^2(2x) \cdot 2 \]
This simplifies to:
\[ \frac{du}{dx} = 2\sec(2x)\tan(2x) + 2\sec^2(2x) \]

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

### Step 5: Factor out the common terms
Factoring out the common factor of 2:
\[ \frac{dy}{dx} = \frac{2(\sec(2x)\tan(2x) + \sec^2(2x))}{\sec(2x) + \tan(2x)} \]

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

---