\( \lim _ { \Delta x \rightarrow 0 } \frac { \sqrt { \tan ( x + \Delta x ) } - \sqrt { \tan x } } { \Delta x } \)

To evaluate the limit \[ \lim _ { \Delta x \rightarrow 0 } \frac { \sqrt { \tan ( x + \Delta x ) } - \sqrt { \tan x } } { \Delta x }, \] we can recognize that this expression resembles the definition of the derivative of a function. Specifically, we can define the function \[ f(x) = \sqrt{\tan x}. \] Then, the limit can be rewritten as \[ \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} = f'(x). \] To find \( f'(x) \), we will use the chain rule. First, we need to find the derivative of \( f(x) = \sqrt{\tan x} \). Using the chain rule, we have: \[ f'(x) = \frac{1}{2\sqrt{\tan x}} \cdot \sec^2 x, \] where \( \sec^2 x \) is the derivative of \( \tan x \). Thus, we can express \( f'(x) \) as: \[ f'(x) = \frac{\sec^2 x}{2\sqrt{\tan x}}. \] Now, substituting this back into our limit, we find: \[ \lim _ { \Delta x \rightarrow 0 } \frac { \sqrt { \tan ( x + \Delta x ) } - \sqrt { \tan x } } { \Delta x } = \frac{\sec^2 x}{2\sqrt{\tan x}}. \] Therefore, the final result is: \[ \boxed{\frac{\sec^2 x}{2\sqrt{\tan x}}}. \]