If `y=log[tan(pi/4 +x/2)])`,then `(dy)/(dx)` is

Options

A. `-tan x`
B. `-sec x`
C. tan x
D. sec x

Answer

To find the derivative of the function \( y = \log(\tan(\frac{\pi}{4} + \frac{x}{2})) \), we will use the chain rule and the properties of logarithms and trigonometric functions.

### Step-by-step Solution:

1. **Identify the function**:
\[ y = \log(\tan(\frac{\pi}{4} + \frac{x}{2})) \]

2. **Differentiate using the chain rule**:
The derivative of \( \log(u) \) is \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \tan(\frac{\pi}{4} + \frac{x}{2}) \).
\[ \frac{dy}{dx} = \frac{1}{\tan(\frac{\pi}{4} + \frac{x}{2})} \cdot \frac{d}{dx}(\tan(\frac{\pi}{4} + \frac{x}{2})) \]

3. **Differentiate \( u = \tan(\frac{\pi}{4} + \frac{x}{2}) \)**:
The derivative of \( \tan(v) \) is \( \sec^2(v) \cdot \frac{dv}{dx} \). Here, \( v = \frac{\pi}{4} + \frac{x}{2} \).
\[ \frac{d}{dx}(\tan(v)) = \sec^2(v) \cdot \frac{d}{dx}(\frac{\pi}{4} + \frac{x}{2}) = \sec^2(\frac{\pi}{4} + \frac{x}{2}) \cdot \frac{1}{2} \]

4. **Substituting back**:
Now substituting back into the derivative:
\[ \frac{dy}{dx} = \frac{1}{\tan(\frac{\pi}{4} + \frac{x}{2})} \cdot \left(\sec^2(\frac{\pi}{4} + \frac{x}{2}) \cdot \frac{1}{2}\right) \]

5. **Using the identity**:
Recall that \( \sec^2(v) = 1 + \tan^2(v) \). Therefore:
\[ \frac{dy}{dx} = \frac{1}{\tan(\frac{\pi}{4} + \frac{x}{2})} \cdot \left(\left(1 + \tan^2(\frac{\pi}{4} + \frac{x}{2})\right) \cdot \frac{1}{2}\right) \]

6. **Simplifying**:
\[ \frac{dy}{dx} = \frac{1}{\tan(\frac{\pi}{4} + \frac{x}{2})} \cdot \frac{1}{2} \cdot \left(1 + \tan^2(\frac{\pi}{4} + \frac{x}{2})\right) \]
\[ = \frac{1}{2} \cdot \frac{1 + \tan^2(\frac{\pi}{4} + \frac{x}{2})}{\tan(\frac{\pi}{4} + \frac{x}{2})} \]

7. **Final Result**:
Thus, the derivative \( \frac{dy}{dx} \) can be expressed as:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sin(\frac{\pi}{4} + \frac{x}{2}) \cdot \cos(\frac{\pi}{4} + \frac{x}{2})} \]