The derivative of `tan^(-1) ((sinx -cosx)/(sinx +cosx))`, with respect to `(x)/(2)`, where `x in(0,(pi)/(2))` is:

Options

A. 1
B. `(1)/(2)`
C. `(1)/(3)`
D. 2

Answer

To find the derivative of \( \tan^{-1} \left( \frac{\sin x - \cos x}{\sin x + \cos x} \right) \) with respect to \( \frac{x}{2} \), we will follow these steps:

### Step 1: Rewrite the expression
We start with the expression:
\[ y = \tan^{-1} \left( \frac{\sin x - \cos x}{\sin x + \cos x} \right) \]

### Step 2: Differentiate with respect to \( x \)
We need to find \( \frac{dy}{d(\frac{x}{2})} \). We can use the chain rule:
\[ \frac{dy}{d(\frac{x}{2})} = \frac{dy}{dx} \cdot \frac{dx}{d(\frac{x}{2})} \]
Since \( \frac{dx}{d(\frac{x}{2})} = 2 \), we have:
\[ \frac{dy}{d(\frac{x}{2})} = 2 \cdot \frac{dy}{dx} \]

### Step 3: Differentiate \( y \) with respect to \( x \)
To differentiate \( y \), we apply the derivative of the inverse tangent function:
\[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{\sin x - \cos x}{\sin x + \cos x} \right)^2} \cdot \frac{d}{dx} \left( \frac{\sin x - \cos x}{\sin x + \cos x} \right) \]

### Step 4: Differentiate the fraction
Using the quotient rule:
\[ \frac{d}{dx} \left( \frac{\sin x - \cos x}{\sin x + \cos x} \right) = \frac{(\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x)}{(\sin x + \cos x)^2} \]

### Step 5: Simplify the derivative
Now, simplifying the numerator:
\[ (\cos x + \sin x)(\sin x + \cos x) - (\sin x - \cos x)(\cos x - \sin x) = (\sin^2 x + \cos^2 x + 2\sin x \cos x) - (-\sin^2 x + \cos^2 x) \]
Using \( \sin^2 x + \cos^2 x = 1 \):
\[ = 1 + 2\sin x \cos x + \sin^2 x - \cos^2 x = 1 + 2\sin x \cos x + 2\cos^2 x \]

### Step 6: Substitute back into the derivative
Now substitute back into the derivative expression:
\[ \frac{dy}{dx} = \frac{1}{1 + \left( \frac{\sin x - \cos x}{\sin x + \cos x} \right)^2} \cdot \frac{1 + 2\sin x \cos x + 2\cos^2 x}{(\sin x + \cos x)^2} \]

### Step 7: Final expression for \( \frac{dy}{d(\frac{x}{2})} \)
Finally, substituting back:
\[ \frac{dy}{d(\frac{x}{2})} = 2 \cdot \frac{1}{1 + \left( \frac{\sin x - \cos x}{\sin x + \cos x} \right)^2} \cdot \frac{1 + 2\sin x \cos x + 2\cos^2 x}{(\sin x + \cos x)^2} \]

### Conclusion
After simplification, we find that the final value of the derivative is \( 2 \).