`((d)/(dx))[log(sec x - tan x)] ` is equal to

Options

A. `-sec x`
B. `sec x + tan x`
C. `sec x `
D. `sec x - tan x `

Answer

To solve the problem \(\frac{d}{dx}[\log(\sec x - \tan x)]\), we will follow these steps:

### Step 1: Set up the function
Let \(y = \log(\sec x - \tan x)\).

### Step 2: Differentiate using the chain rule
Using the chain rule, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{1}{\sec x - \tan x} \cdot \frac{d}{dx}(\sec x - \tan x) \]

### Step 3: Differentiate \(\sec x\) and \(\tan x\)
Now we need to differentiate \(\sec x\) and \(\tan x\):
- The derivative of \(\sec x\) is \(\sec x \tan x\).
- The derivative of \(\tan x\) is \(\sec^2 x\).

So, we have:
\[ \frac{d}{dx}(\sec x - \tan x) = \sec x \tan x - \sec^2 x \]

### Step 4: Substitute back into the derivative
Substituting this back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{\sec x - \tan x} \cdot (\sec x \tan x - \sec^2 x) \]

### Step 5: Factor the numerator
We can factor the numerator:
\[ \sec x \tan x - \sec^2 x = \sec x (\tan x - \sec x) \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{\sec x (\tan x - \sec x)}{\sec x - \tan x} \]

### Step 6: Simplify the expression
Now, we can simplify:
\[ \frac{dy}{dx} = \frac{\sec x (\tan x - \sec x)}{\sec x - \tan x} \]
Notice that \(\tan x - \sec x = -(\sec x - \tan x)\), so we can rewrite:
\[ \frac{dy}{dx} = \frac{\sec x (-1)(\sec x - \tan x)}{\sec x - \tan x} \]
The \((\sec x - \tan x)\) terms cancel out:
\[ \frac{dy}{dx} = -\sec x \]

### Final Answer
Thus, we conclude that:
\[ \frac{d}{dx}[\log(\sec x - \tan x)] = -\sec x \]

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