The given function is \( \displaystyle y = \sqrt{\frac{\sec x \,- 1}{\sec x + 1}} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
First, simplify the expression inside the square root:
\( \begin{aligned} y&= \sqrt{\frac{\sec x \,- 1}{\sec x + 1}} \\ &= \sqrt{\frac{\sec x \,- 1}{\sec x + 1} \cdot \frac{\cos x}{\cos x}} \\ &= \sqrt{\frac{1 \,- \cos x}{1 + \cos x}} = \sqrt{\tan^2\left(\frac{x}{2}\right)} \\ &= \left|\tan\left(\frac{x}{2}\right)\right| \\ \end{aligned} \)Assuming \(\displaystyle \tan\left(\frac{x}{2}\right) \geq 0\) (for simplicity in differentiation), we have \(\displaystyle y = \tan\left(\frac{x}{2}\right) \).
Now, differentiate \(y\) with respect to \(x\):
\( \begin{aligned} \frac{dy}{dx} &= \frac{d}{dx} \left[\tan\left(\frac{x}{2}\right)\right] \\ &= \frac{1}{2} \sec^2\left(\frac{x}{2}\right) \\ &= \frac{1}{2\cos^2\left(\frac{x}{2}\right)} \\ &= \frac{1}{1 + \cos x} \quad \because 2\cos^2\left(\frac{x}{2}\right) = 1 + \cos x \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = \frac{1}{1 + \cos x}} \\ \)Please let me know in the comments if you find any errors in this solution.