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.)

Since \[ f(x)=1+\sqrt{5x}, \] we begin with the definition of the derivative: \[ f^{\prime}(x)=\lim_{z \rightarrow x}\frac{f(z)-f(x)}{z-x}. \] 1. Substitute \( f(z)=1+\sqrt{5z} \) and \( f(x)=1+\sqrt{5x} \): \[ f^{\prime}(x)=\lim_{z \rightarrow x}\frac{(1+\sqrt{5z})-(1+\sqrt{5x})}{z-x} = \lim_{z \rightarrow x}\frac{\sqrt{5z}-\sqrt{5x}}{z-x}. \] 2. Multiply the numerator and denominator by the conjugate \(\sqrt{5z}+\sqrt{5x}\): \[ \frac{\sqrt{5z}-\sqrt{5x}}{z-x} \cdot \frac{\sqrt{5z}+\sqrt{5x}}{\sqrt{5z}+\sqrt{5x}} = \frac{(\sqrt{5z}-\sqrt{5x})(\sqrt{5z}+\sqrt{5x})}{(z-x)(\sqrt{5z}+\sqrt{5x})}. \] 3. Simplify the numerator using the difference of squares: \[ (\sqrt{5z}-\sqrt{5x})(\sqrt{5z}+\sqrt{5x}) = (5z-5x). \] Thus, \[ \frac{5z-5x}{(z-x)(\sqrt{5z}+\sqrt{5x})} = \frac{5(z-x)}{(z-x)(\sqrt{5z}+\sqrt{5x})}. \] 4. Cancel the common factor \((z-x)\) (for \( z \neq x \)): \[ = \frac{5}{\sqrt{5z}+\sqrt{5x}}. \] 5. Take the limit as \( z \rightarrow x \): \[ f^{\prime}(x)=\lim_{z \rightarrow x}\frac{5}{\sqrt{5z}+\sqrt{5x}} = \frac{5}{2\sqrt{5x}}. \] 6. Simplify the expression: \[ \frac{5}{2\sqrt{5x}} = \frac{5}{2\sqrt{5}\sqrt{x}} = \frac{\sqrt{5}}{2\sqrt{x}}. \] Thus, the derivative is: \[ f^{\prime}(x)=\frac{\sqrt{5}}{2\sqrt{x}}. \]