`(d)/(dx)[log(secx-tanx)]` is equal to

Options

A. `-secx`
B. `secx+tanx`
C. `secx`
D. `secx-tanx`

Answer

To find the derivative of the function \( y = \log(\sec x - \tan x) \), we will use the chain rule and the derivatives of the logarithmic, secant, and tangent functions.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = \log(\sec x - \tan x) \]

2. **Differentiate using the chain rule**:
The derivative of \( \log(u) \) with respect to \( x \) is given by:
\[ \frac{dy}{dx} = \frac{1}{u} \cdot \frac{du}{dx} \]
where \( u = \sec x - \tan x \).

3. **Find \( \frac{du}{dx} \)**:
We need to differentiate \( u = \sec x - \tan x \).
- The derivative of \( \sec x \) is \( \sec x \tan x \).
- The derivative of \( \tan x \) is \( \sec^2 x \).
Thus,
\[ \frac{du}{dx} = \sec x \tan x - \sec^2 x \]

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

5. **Simplify the expression**:
We can rewrite the expression:
\[ \frac{dy}{dx} = \frac{\sec x \tan x - \sec^2 x}{\sec x - \tan x} \]

6. **Factor out common terms**:
Notice that we can factor out \( \sec x \) from the numerator:
\[ \frac{dy}{dx} = \frac{\sec x (\tan x - \sec x)}{\sec x - \tan x} \]

7. **Final simplification**:
The numerator and denominator can be rearranged:
\[ \frac{dy}{dx} = -\sec x \]

### Final Result:
Thus, the derivative \( \frac{d}{dx}[\log(\sec x - \tan x)] \) is:
\[ \frac{dy}{dx} = -\sec x \]