Differentiate `sqrt((1-tanx)/(1+tanx))` w.r.t.x.

Answer

To differentiate the function \( y = \sqrt{\frac{1 - \tan x}{1 + \tan x}} \) with respect to \( x \), we will follow these steps:

### Step 1: Let \( t = \frac{1 - \tan x}{1 + \tan x} \)
We start by letting \( t \) represent the expression inside the square root.

### Step 2: Rewrite \( y \) in terms of \( t \)
Now we can express \( y \) in terms of \( t \):
\[ y = \sqrt{t} \]

### Step 3: Differentiate \( y \) with respect to \( x \)
Using the chain rule, we differentiate \( y \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{t}} \cdot \frac{dt}{dx} \]

### Step 4: Differentiate \( t \) with respect to \( x \)
To find \( \frac{dt}{dx} \), we will use the quotient rule on \( t = \frac{1 - \tan x}{1 + \tan x} \):
Let \( u = 1 - \tan x \) and \( v = 1 + \tan x \).
Using the quotient rule:
\[ \frac{dt}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} \]
where:
- \( \frac{du}{dx} = -\sec^2 x \) (derivative of \( 1 - \tan x \))
- \( \frac{dv}{dx} = \sec^2 x \) (derivative of \( 1 + \tan x \))

Substituting these into the quotient rule:
\[ \frac{dt}{dx} = \frac{(1 + \tan x)(-\sec^2 x) - (1 - \tan x)(\sec^2 x)}{(1 + \tan x)^2} \]

### Step 5: Simplify \( \frac{dt}{dx} \)
Expanding the numerator:
\[ = \frac{-\sec^2 x - \tan x \sec^2 x - \sec^2 x + \tan x \sec^2 x}{(1 + \tan x)^2} \]
This simplifies to:
\[ = \frac{-2\sec^2 x}{(1 + \tan x)^2} \]

### Step 6: Substitute \( \frac{dt}{dx} \) back into \( \frac{dy}{dx} \)
Now we substitute \( \frac{dt}{dx} \) back into the expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{t}} \cdot \left(-\frac{2\sec^2 x}{(1 + \tan x)^2}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = -\frac{\sec^2 x}{\sqrt{t} (1 + \tan x)^2} \]

### Step 7: Substitute \( t \) back into the expression
Recall that \( t = \frac{1 - \tan x}{1 + \tan x} \):
\[ \sqrt{t} = \sqrt{\frac{1 - \tan x}{1 + \tan x}} \]
Thus, the final expression for \( \frac{dy}{dx} \) becomes:
\[ \frac{dy}{dx} = -\frac{\sec^2 x}{\sqrt{\frac{1 - \tan x}{1 + \tan x}} (1 + \tan x)^2} \]

### Final Result
\[ \frac{dy}{dx} = -\frac{\sec^2 x}{(1 + \tan x)^2 \sqrt{\frac{1 - \tan x}{1 + \tan x}}} \]