If ` y= log (secx+tan x ) ,then (dy)/(dx) `

Options

A. ` secx `
B. `secx tan x `
C. ` tanx`
D. ` cos x `

Answer

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

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

### Step 2: Apply the chain rule for differentiation
The derivative of \( \log(f(x)) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{f(x)} \cdot f'(x) \]
where \( f(x) = \sec x + \tan x \).

### Step 3: Differentiate \( f(x) \)
Now, we need to find \( f'(x) \):
\[ f(x) = \sec x + \tan x \]
The derivatives of \( \sec x \) and \( \tan x \) are:
\[ \frac{d}{dx}(\sec x) = \sec x \tan x \]
\[ \frac{d}{dx}(\tan x) = \sec^2 x \]
Thus,
\[ f'(x) = \sec x \tan x + \sec^2 x \]

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

### Step 5: Simplify the expression
We can factor out \( \sec x \) from the numerator:
\[ \frac{dy}{dx} = \frac{\sec x (\tan x + \sec x)}{\sec x + \tan x} \]
Now, notice that \( \sec x + \tan x \) in the numerator and denominator can be cancelled:
\[ \frac{dy}{dx} = \sec x \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = \sec x \]

---