Use the formula \( f^{\prime}(x)=\lim _{z \rightarrow x} \frac{f(z)-f(x)}{z-x} \) to find the derivative of the following function. \( f(x)=1+\sqrt{5 x} \) \( f^{\prime}(x)=\square \) (Type an exact answer, using radicals as needed.)

To find the derivative of the function \( f(x) = 1 + \sqrt{5x} \) using the given formula, we start by substituting \( f(z) \) and \( f(x) \) into the limit. 1. Calculate \( f(z) \): \[ f(z) = 1 + \sqrt{5z} \] 2. Set up the limit: \[ f^{\prime}(x) = \lim_{z \rightarrow x} \frac{f(z) - f(x)}{z - x} = \lim_{z \rightarrow x} \frac{(1 + \sqrt{5z}) - (1 + \sqrt{5x})}{z - x} \] 3. Simplify the expression: \[ f^{\prime}(x) = \lim_{z \rightarrow x} \frac{\sqrt{5z} - \sqrt{5x}}{z - x} \] 4. We can factor the numerator using the difference of squares: \[ \sqrt{5z} - \sqrt{5x} = \frac{5z - 5x}{\sqrt{5z} + \sqrt{5x}} = \frac{5(z - x)}{\sqrt{5z} + \sqrt{5x}} \] 5. Substitute back into the limit: \[ f^{\prime}(x) = \lim_{z \rightarrow x} \frac{5(z - x)}{(z - x)(\sqrt{5z} + \sqrt{5x})} \] 6. Cancel \( z - x \) (as \( z \neq x \) in the limit): \[ f^{\prime}(x) = \lim_{z \rightarrow x} \frac{5}{\sqrt{5z} + \sqrt{5x}} \] 7. Now substitute \( z = x \): \[ f^{\prime}(x) = \frac{5}{\sqrt{5x} + \sqrt{5x}} = \frac{5}{2\sqrt{5x}} = \frac{5}{2\sqrt{5} \sqrt{x}} = \frac{\sqrt{5}}{2x} \] Thus, the derivative of the function is: \[ f^{\prime}(x) = \frac{\sqrt{5}}{2x} \]