Find the derivative of the following function: $\frac{\sec x-1}{\sec x+1}$

Let $f(x) = \frac{\sec x - 1}{\sec x + 1}$.
We can simplify the function using trigonometric identities:
$f(x) = \frac{\frac{1}{\cos x} - 1}{\frac{1}{\cos x} + 1} = \frac{1 - \cos x}{1 + \cos x} = \frac{2 \sin^2(x/2)}{2 \cos^2(x/2)} = \tan^2(x/2)$.
Now,differentiate $f(x) = \tan^2(x/2)$ with respect to $x$ using the chain rule:
$f'(x) = 2 \tan(x/2) \cdot \sec^2(x/2) \cdot \frac{1}{2} = \tan(x/2) \sec^2(x/2)$.
Alternatively,using the quotient rule on $\frac{1 - \cos x}{1 + \cos x}$:
$f'(x) = \frac{(1 + \cos x)(\sin x) - (1 - \cos x)(-\sin x)}{(1 + \cos x)^2}$
$f'(x) = \frac{\sin x + \sin x \cos x + \sin x - \sin x \cos x}{(1 + \cos x)^2}$
$f'(x) = \frac{2 \sin x}{(1 + \cos x)^2}$.