If `y=(secx-tanx)/(secx+tanx),` then `(dy)/(dx)` equals.

Options

A. `2secx(secx-tanx)`
B. `-2secx(secx-tanx)^(2)`
C. `2secx(secx-tanx)^(2)`
D. `-2secx(secx+tanx)^(2)`

Answer

To find the derivative of the function \( y = \frac{\sec x - \tan x}{\sec x + \tan x} \), we will follow these steps:

### Step 1: Rationalize the expression
We can rationalize the expression by multiplying the numerator and denominator by the conjugate of the denominator:

\[ y = \frac{(\sec x - \tan x)(\sec x - \tan x)}{(\sec x + \tan x)(\sec x - \tan x)} \]

### Step 2: Simplify the numerator and denominator
Using the identity \( a^2 - b^2 = (a + b)(a - b) \):

- The numerator becomes:
\[ (\sec x - \tan x)^2 = \sec^2 x - 2\sec x \tan x + \tan^2 x \]

- The denominator simplifies using the identity \( \sec^2 x - \tan^2 x = 1 \):
\[ (\sec x + \tan x)(\sec x - \tan x) = \sec^2 x - \tan^2 x = 1 \]

Thus, we have:
\[ y = \sec^2 x - 2\sec x \tan x \]

### Step 3: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \frac{d}{dx}(\sec^2 x - 2\sec x \tan x) \]

Using the product rule and chain rule, we differentiate each term:

1. The derivative of \( \sec^2 x \) is \( 2\sec^2 x \tan x \).
2. The derivative of \( -2\sec x \tan x \) using the product rule:
\[ -2(\sec x \cdot \frac{d}{dx}(\tan x) + \tan x \cdot \frac{d}{dx}(\sec x)) = -2(\sec x \sec^2 x + \tan x \sec x \tan x) \]

Combining these results, we have:
\[ \frac{dy}{dx} = 2\sec^2 x \tan x - 2(\sec^3 x + \sec x \tan^2 x) \]

### Step 4: Factor out common terms
Factoring out \( 2\sec x \):
\[ \frac{dy}{dx} = 2\sec x (\sec x \tan x - \sec^2 x - \tan^2 x) \]

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