If `f(x) =((sec x - 1)/( sec x + 1))^(1/2)` , find `f'((4pi)/( 3))`

Answer

To find \( f'\left(\frac{4\pi}{3}\right) \) for the function \( f(x) = \left(\frac{\sec x - 1}{\sec x + 1}\right)^{\frac{1}{2}} \), we will follow these steps:

### Step 1: Rewrite the function
We start with the function:
\[ f(x) = \left(\frac{\sec x - 1}{\sec x + 1}\right)^{\frac{1}{2}} \]
Using the identity \( \sec x = \frac{1}{\cos x} \), we can rewrite \( f(x) \):
\[ f(x) = \left(\frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1}\right)^{\frac{1}{2}} \]

### Step 2: Simplify the expression
Now, we simplify the fraction:
\[ f(x) = \left(\frac{\frac{1 - \cos x}{\cos x}}{\frac{1 + \cos x}{\cos x}}\right)^{\frac{1}{2}} = \left(\frac{1 - \cos x}{1 + \cos x}\right)^{\frac{1}{2}} \]

### Step 3: Rationalize the expression
Next, we rationalize the expression:
\[ f(x) = \sqrt{\frac{1 - \cos x}{1 + \cos x}} = \frac{\sqrt{1 - \cos x}}{\sqrt{1 + \cos x}} \]

### Step 4: Use trigonometric identities
Using the identity \( 1 - \cos^2 x = \sin^2 x \), we can express \( \sqrt{1 - \cos x} \) as:
\[ f(x) = \frac{\sqrt{1 - \cos^2 x}}{\sqrt{1 + \cos x}} = \frac{\sin x}{\sqrt{1 + \cos x}} \]

### Step 5: Differentiate the function
Now we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(\frac{\sin x}{\sqrt{1 + \cos x}}\right) \]
Using the quotient rule:
\[ f'(x) = \frac{\sqrt{1 + \cos x} \cdot \cos x - \sin x \cdot \frac{-\sin x}{2\sqrt{1 + \cos x}}}{(1 + \cos x)} \]
This simplifies to:
\[ f'(x) = \frac{\cos x \sqrt{1 + \cos x} + \frac{\sin^2 x}{2\sqrt{1 + \cos x}}}{1 + \cos x} \]

### Step 6: Evaluate at \( x = \frac{4\pi}{3} \)
Now we substitute \( x = \frac{4\pi}{3} \):
\[ f'\left(\frac{4\pi}{3}\right) = \frac{\cos\left(\frac{4\pi}{3}\right) \sqrt{1 + \cos\left(\frac{4\pi}{3}\right)} + \frac{\sin^2\left(\frac{4\pi}{3}\right)}{2\sqrt{1 + \cos\left(\frac{4\pi}{3}\right)}}}{1 + \cos\left(\frac{4\pi}{3}\right)} \]
Calculating \( \cos\left(\frac{4\pi}{3}\right) = -\frac{1}{2} \) and \( \sin\left(\frac{4\pi}{3}\right) = -\frac{\sqrt{3}}{2} \):
\[ 1 + \cos\left(\frac{4\pi}{3}\right) = 1 - \frac{1}{2} = \frac{1}{2} \]
Thus:
\[ f'\left(\frac{4\pi}{3}\right) = \frac{-\frac{1}{2} \sqrt{\frac{1}{2}} + \frac{\left(-\frac{\sqrt{3}}{2}\right)^2}{2\sqrt{\frac{1}{2}}}}{\frac{1}{2}} \]
Calculating \( \sin^2\left(\frac{4\pi}{3}\right) = \frac{3}{4} \):
\[ f'\left(\frac{4\pi}{3}\right) = \frac{-\frac{1}{2} \cdot \frac{1}{\sqrt{2}} + \frac{\frac{3}{4}}{2\cdot\frac{1}{\sqrt{2}}}}{\frac{1}{2}} = 2 \]

### Final Answer
Thus, the value of \( f'\left(\frac{4\pi}{3}\right) \) is:
\[ \boxed{2} \]