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

Options

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

Answer

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 start with the given function:
\[ y = \sqrt{\frac{1 + \tan x}{1 - \tan x}} \]

### Step 2: Apply the chain rule
To differentiate \( y \), we will use the chain rule. The derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \) where \( u = \frac{1 + \tan x}{1 - \tan x} \).

Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{\frac{1 + \tan x}{1 - \tan x}}} \cdot \frac{d}{dx}\left(\frac{1 + \tan x}{1 - \tan x}\right) \]

### Step 3: Differentiate the inner function
Next, we need to differentiate \( \frac{1 + \tan x}{1 - \tan x} \). We will use the quotient rule, which 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 \( \frac{du}{dx} \) and \( \frac{dv}{dx} \):
- \( \frac{du}{dx} = \sec^2 x \)
- \( \frac{dv}{dx} = -\sec^2 x \)

Now applying 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:
\[ \frac{(1 - \tan x)\sec^2 x + (1 + \tan x)\sec^2 x}{(1 - \tan x)^2} = \frac{(1 - \tan x + 1 + \tan x)\sec^2 x}{(1 - \tan x)^2} = \frac{2\sec^2 x}{(1 - \tan x)^2} \]

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

### Step 6: Simplify the final expression
The \( 2 \) in the numerator and denominator cancels out:
\[ \frac{dy}{dx} = \frac{\sec^2 x}{\sqrt{\frac{1 + \tan x}{1 - \tan x}}(1 - \tan x)^2} \]

### Step 7: Further simplify the square root
We can express the square root as:
\[ \sqrt{\frac{1 + \tan x}{1 - \tan x}} = \frac{\sqrt{1 + \tan x}}{\sqrt{1 - \tan x}} \]
Thus, we can rewrite \( \frac{dy}{dx} \) as:
\[ \frac{dy}{dx} = \frac{\sec^2 x \cdot \sqrt{1 - \tan x}}{(1 - \tan x)^2 \cdot \sqrt{1 + \tan x}} \]

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