`(d)/(dx)((secx+tanx)/(secx-tanx))=`

Options

A. `(2*cosx)/((1-sinx)^(2))`
B. `(cosx)/((1-sinx)^(2))`
C. `(2*cosx)/((1-sinx))`
D. none of theses

Answer

To solve the problem \(\frac{d}{dx}\left(\frac{\sec x + \tan x}{\sec x - \tan x}\right)\), we will use the quotient rule and some trigonometric identities. Here’s a step-by-step solution:

### Step 1: Identify the function
Let \(y = \frac{\sec x + \tan x}{\sec x - \tan x}\).

### Step 2: Apply the Quotient Rule
The quotient rule states that if you have a function \(\frac{u}{v}\), then the derivative is given by:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
Here, \(u = \sec x + \tan x\) and \(v = \sec x - \tan x\).

### Step 3: Differentiate \(u\) and \(v\)
First, we need to find \(\frac{du}{dx}\) and \(\frac{dv}{dx}\).

1. **Differentiate \(u\)**:
\[ \frac{du}{dx} = \frac{d}{dx}(\sec x) + \frac{d}{dx}(\tan x) = \sec x \tan x + \sec^2 x \]

2. **Differentiate \(v\)**:
\[ \frac{dv}{dx} = \frac{d}{dx}(\sec x) - \frac{d}{dx}(\tan x) = \sec x \tan x - \sec^2 x \]

### Step 4: Substitute into the Quotient Rule
Now substitute \(u\), \(v\), \(\frac{du}{dx}\), and \(\frac{dv}{dx}\) into the quotient rule:
\[ \frac{d}{dx}\left(\frac{\sec x + \tan x}{\sec x - \tan x}\right) = \frac{(\sec x - \tan x)(\sec x \tan x + \sec^2 x) - (\sec x + \tan x)(\sec x \tan x - \sec^2 x)}{(\sec x - \tan x)^2} \]

### Step 5: Simplify the numerator
Let’s simplify the numerator:
1. Expand both products:
\[ (\sec x - \tan x)(\sec x \tan x + \sec^2 x) = \sec x \tan x \sec x - \tan x \sec x \tan x + \sec^2 x \sec x - \tan x \sec^2 x \]
\[ = \sec^2 x \tan x - \tan^2 x \sec x + \sec^3 x - \tan x \sec^2 x \]

2. For the second part:
\[ (\sec x + \tan x)(\sec x \tan x - \sec^2 x) = \sec x \tan x \sec x + \tan x \sec x \tan x - \sec^2 x \sec x - \tan x \sec^2 x \]
\[ = \sec^2 x \tan x + \tan^2 x \sec x - \sec^3 x - \tan x \sec^2 x \]

3. Combine these results:
\[ \text{Numerator} = \sec^2 x \tan x - \tan^2 x \sec x + \sec^3 x - \tan x \sec^2 x - (\sec^2 x \tan x + \tan^2 x \sec x - \sec^3 x - \tan x \sec^2 x) \]
\[ = 2\sec^3 x - 2\tan^2 x \sec x \]

### Step 6: Write the final derivative
Thus, the derivative is:
\[ \frac{d}{dx}\left(\frac{\sec x + \tan x}{\sec x - \tan x}\right) = \frac{2(\sec^3 x - \tan^2 x \sec x)}{(\sec x - \tan x)^2} \]