log (sec 2x +tan 2 x)

Answer

To differentiate the function \( y = \log(\sec(2x) + \tan(2x)) \) with respect to \( x \), we will follow these steps:

### Step 1: Differentiate the logarithmic function
Using the chain rule, the derivative of \( y = \log(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = \sec(2x) + \tan(2x) \).

### Step 2: Differentiate \( u \)
Now we need to find \( \frac{du}{dx} \):
\[ u = \sec(2x) + \tan(2x) \]
The derivative of \( \sec(2x) \) is \( 2\sec(2x)\tan(2x) \) and the derivative of \( \tan(2x) \) is \( 2\sec^2(2x) \). Therefore:
\[ \frac{du}{dx} = 2\sec(2x)\tan(2x) + 2\sec^2(2x) \]

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

### Step 4: Factor out common terms
We can factor out the common term \( 2 \):
\[ \frac{dy}{dx} = \frac{2(\sec(2x)\tan(2x) + \sec^2(2x))}{\sec(2x) + \tan(2x)} \]

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