If `y = sqrt((1 + tan x)/(1 - tan x)) " then " (dy)/(dx)=` ?

Options

A. `(1)/(2) sec^(2)x.tan (x + (pi)/(4))`
B. `(sec^(2) (x + (pi)/(4)))/(2 sqrt(tan (x + (pi)/(4))))`
C. `(sec^(2) ((x)/(4)))/(sqrt(tan (x + (pi)/(4))))`
D. none of these

Explanation

To find the derivative \(\frac{dy}{dx}\) for the function \(y = \sqrt{\frac{1 + \tan x}{1 - \tan x}}\), we will follow these steps:

### Step 1: Rewrite the function
We can rewrite the function as:
\[ y = \left(\frac{1 + \tan x}{1 - \tan x}\right)^{1/2} \]

### Step 2: Apply the chain rule
Using the chain rule for differentiation, we have:
\[ \frac{dy}{dx} = \frac{1}{2} \left(\frac{1 + \tan x}{1 - \tan x}\right)^{-1/2} \cdot \frac{d}{dx}\left(\frac{1 + \tan x}{1 - \tan x}\right) \]

### Step 3: Differentiate the inner function
Now, we need to differentiate the inner function \(\frac{1 + \tan x}{1 - \tan x}\) using the quotient rule. The quotient rule states that if \(u = 1 + \tan x\) and \(v = 1 - \tan x\), then:
\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2} \]

Calculating \(du\) and \(dv\):
- \(du = \sec^2 x \, dx\)
- \(dv = -\sec^2 x \, dx\)

Now substituting into the quotient rule:
\[ \frac{d}{dx}\left(\frac{1 + \tan x}{1 - \tan x}\right) = \frac{(1 - \tan x)(\sec^2 x) - (1 + \tan x)(-\sec^2 x)}{(1 - \tan x)^2} \]

### Step 4: Simplify the derivative
This simplifies to:
\[ \frac{(1 - \tan x + 1 + \tan x)\sec^2 x}{(1 - \tan x)^2} = \frac{2\sec^2 x}{(1 - \tan x)^2} \]

### Step 5: Substitute back into the derivative
Now substituting back into our expression for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{1}{2} \left(\frac{1 + \tan x}{1 - \tan x}\right)^{-1/2} \cdot \frac{2\sec^2 x}{(1 - \tan x)^2} \]

### Step 6: Final expression
This simplifies to:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{(1 - \tan x)^2 \sqrt{\frac{1 + \tan x}{1 - \tan x}}} \]

### Final Answer
Thus, the derivative is:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{(1 - \tan x)^2 \sqrt{\frac{1 + \tan x}{1 - \tan x}}} \]